university of Žilina · obsahzoznam01112131415161718191 zoznamintegrálov–príklady101–200...

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91

Matematická analýza 1

2018/2019

13. Neurčitý integrálRiešené príklady

[email protected] http://frcatel.fri.uniza.sk/˜beerb

mailto:[email protected]://frcatel.fri.uniza.sk/~beerb

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91

Obsah – príklady 101–200

1 Riešené príklady 101–1102 Riešené príklady 111–1203 Riešené príklady 121–1304 Riešené príklady 131–1405 Riešené príklady 141–1506 Riešené príklady 151–1607 Riešené príklady 161–1708 Riešené príklady 171–1809 Riešené príklady 181–19010 Riešené príklady 191–200

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91

Zoznam integrálov – príklady 101–200101.

∫dx

x2+4x+5 102.

∫dx

x2+4x+3 103.

∫dx

x2−4x+6 104.

∫dx

x2−4x+2 105.

∫dx

x2+4x+4 106.

∫dx√

x2+4x+4107.

∫dx√

x2+4x+5108.

∫dx√

x2+4x+3

109.

∫dx√

x2−4x+6110.

∫dx√

x2−4x+2111.

∫dx√

−x2+4x−5112.

∫dx√

−x2+4x−3113.

∫dx√

x(x−1)114.

∫dx√

x(2−x)115.

∫√

x2+4x +5 dx

116.

∫√

x2+4x +3 dx 117.∫√

x2−4x +6 dx 118.∫√

x2−4x +2 dx 119.∫√−x2+4x−3 dx 120.

∫ √x(2−x) dx 121.

∫dx

(x2+a2)n 122.

∫dx

(x2+a2)2

123.

∫dx

(x2+a2)3 124.

∫dx

(x2+a2)4 125.

∫dx

(x2−a2)n 126.

∫dx

(x2−a2)2 127.

∫dx

(x2−a2)3 128.

∫dx

(x2−a2)4 129.

∫dx

(x2+4x+5)2 130.

∫dx

(x2+4x+3)2 131.

∫x2 dx

(x2+a2)2

132.

∫x2 dx

(x2−a2)2 133.

∫x dx

x2+a2 134.

∫x dx

(x2+a2)n 135.

∫x dx

x2−a2 136.

∫x dx

(x2−a2)n 137.

∫xn dxx−1 138.

∫x dxx−1 139.

∫x2 dxx−1 140.

∫x9 dxx−1 141.

∫dx

(1−x)x2

142.

∫dx

x6(1+x2) 143.

∫(x−2)4 dx

(x−1)2 144.

∫(x−1)4 dx

(x−2)2 145.

∫dx

x3−7x−6 146.

∫dx

x3−2x2−x+2 147.

∫dx

x3−3x−2 148.

∫dx

x3+x2−x−1 149.

∫dx

x3−3x2+4x−2

150.

∫dx

x3−3x2+3x−1 151.

∫dx

x3−2x−4 152.

∫dx

x6+1 153.

∫x dxx6+1 154.

∫x2 dxx6+1 155.

∫x3 dxx6+1 156.

∫x4 dxx6+1 157.

∫x5 dxx6+1 158.

∫x6 dxx6+1 159.

∫dx

2x +1 160.

∫dx√2x +1

161.

∫(1+x) dx√

1−x2162.

∫ √1+x1−x dx 163.

∫ √1−x1+x dx 164.

∫ √x+1x−1 dx 165.

∫ √x−1x+1 dx 166.

∫ √(1+x1−x

)3 dx 167.∫ √( 1−x1+x )3 dx168.

∫ √(x+1x−1

)3 dx 169.∫ √( x−1x+1 )3 dx 170.∫ √1−√x√1+√x dx 171.∫ dx√x+1+ 3√x+1 172.∫ arcsin√ 1+x1−x dx 173.∫ arcsin√ 1−x1+x dx174.

∫arcsin

√x+1x−1 dx 175.

∫arcsin

√x−1x+1 dx 176.

∫dx√

x−3+√

x−5 177.

∫dx√

x−3−√

x−5 178.

∫dx√

x−3+√5−x 179.

∫1+√

1−x2

1−√

1−x2dx

180.

∫ √ x1−x√

x dx 181.∫

dx3√x+ 4√

x 182.

∫dx

6√x+ 4√

x 183.

∫dx

3√x+ 5√

x 184.

∫dx

3√x+1 185.

∫ [√x3 − 1√x

]dx 186.

∫x−1

(√

x+ 3√

x2)xdx 187.

∫dx

(x−1)√

x−2

188.

∫dx

(x+1)√1−x 189.

∫(x−1)

√x−2 dx 190.

∫√1−x

x+1 dx 191.∫

dx(x−√

x2−1)2 192.∫ x5 dx√x2+1 193.

∫x5 dx√

x3+1194.

∫x5 dx√x2−1

195.

∫x5 dx√x3−1

196.

∫x5 dx√1−x2

197.

∫x5 dx√1−x3

198.

∫ √1−x2x2 dx 199.

∫dx

x√

x4+x2+1200.

∫dx

x√

x6+x3+1

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

Riešené príklady – 101, 102∫dx

x2+4x+5 = arctg (x +2) + c

=∫

dxx2+4x+4+1 =

∫dx

(x+2)2+1 =[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

dtt2+1 =arctg t + c = arctg (x +2) + c, x ∈R, c∈R.

∫dx

x2+4x+3 =12 ln∣∣ x+1

x+3∣∣+ c

=∫

dxx2+4x+4−1 =

∫dx

(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}

dt =dx t∈R−{±1}

]

=∫

dtt2−1 =

12 ln∣∣ t−1

t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,

x ∈R−{−1,− 3}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = arctg (x +2) + c

=∫

dxx2+4x+4+1 =

∫dx

(x+2)2+1 =[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

∫dx

x2+4x+3 =12 ln∣∣ x+1

x+3∣∣+ c

=∫

dxx2+4x+4−1 =

∫dx

(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}

]

=∫

dtt2−1 =

12 ln∣∣ t−1

t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,

x ∈R−{−1,− 3}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = arctg (x +2) + c

=∫

dxx2+4x+4+1 =

∫dx

(x+2)2+1 =[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

∫dx

x2+4x+3 =12 ln∣∣ x+1

x+3∣∣+ c

=∫

dxx2+4x+4−1 =

∫dx

(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}

]

=∫

dtt2−1 =

12 ln∣∣ t−1

t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,

x ∈R−{−1,− 3}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = arctg (x +2) + c

=∫

dxx2+4x+4+1 =

∫dx

(x+2)2+1 =[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

∫dx

x2+4x+3 =12 ln∣∣ x+1

x+3∣∣+ c

=∫

dxx2+4x+4−1 =

∫dx

(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}

]

=∫

dtt2−1 =

12 ln∣∣ t−1

t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,

x ∈R−{−1,− 3}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = arctg (x +2) + c

=∫

dxx2+4x+4+1 =

∫dx

(x+2)2+1 =[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

∫dx

x2+4x+3 =12 ln∣∣ x+1

x+3∣∣+ c

=∫

dxx2+4x+4−1 =

∫dx

(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}

]

=∫

dtt2−1 =

12 ln∣∣ t−1

t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,

x ∈R−{−1,− 3}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = arctg (x +2) + c

=∫

dxx2+4x+4+1 =

∫dx

(x+2)2+1 =[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

∫dx

x2+4x+3 =12 ln∣∣ x+1

x+3∣∣+ c

=∫

dxx2+4x+4−1 =

∫dx

(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}

]

=∫

dtt2−1 =

12 ln∣∣ t−1

t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,

x ∈R−{−1,− 3}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = arctg (x +2) + c

=∫

dxx2+4x+4+1 =

∫dx

(x+2)2+1 =[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

∫dx

x2+4x+3 =12 ln∣∣ x+1

x+3∣∣+ c

=∫

dxx2+4x+4−1 =

∫dx

(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}

]

=∫

dtt2−1 =

12 ln∣∣ t−1

t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,

x ∈R−{−1,− 3}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6 = arctg (x +2) + c

=∫

dxx2−4x+4+2 =

∫dx

(x−2)2+(√2)2 =

[ Subst. t =x−2 x ∈Rdt =dx t∈R

]

=∫

dtt2+(√2)2 =

1√2 arctg

t√2 + c =

1√2 arctg

x−2√2 + c, x ∈R, c∈R.

∫dx

x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c

=∫

dxx2−4x+4−2 =

∫dx

(x−2)2−(√2)2 =

[Subst. t =x−2 x ∈R−{2±

√2}

dt =dx t∈R−{±√2}

]

=∫

dtt2−(√2)2 =

12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,

x ∈R−{2±√2}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6 = arctg (x +2) + c

=∫

dxx2−4x+4+2 =

∫dx

(x−2)2+(√2)2 =

]

=∫

dtt2+(√2)2 =

1√2 arctg

t√2 + c =

1√2 arctg

x−2√2 + c, x ∈R, c∈R.

∫dx

x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c

=∫

dxx2−4x+4−2 =

∫dx

(x−2)2−(√2)2 =

[Subst. t =x−2 x ∈R−{2±

√2}

dt =dx t∈R−{±√2}

]

=∫

dtt2−(√2)2 =

12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,

x ∈R−{2±√2}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6 = arctg (x +2) + c

=∫

dxx2−4x+4+2 =

∫dx

(x−2)2+(√2)2 =

]

=∫

dtt2+(√2)2 =

1√2 arctg

t√2 + c =

1√2 arctg

x−2√2 + c, x ∈R, c∈R.

∫dx

x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c

=∫

dxx2−4x+4−2 =

∫dx

(x−2)2−(√2)2 =

[Subst. t =x−2 x ∈R−{2±

√2}

dt =dx t∈R−{±√2}

]

=∫

dtt2−(√2)2 =

12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,

x ∈R−{2±√2}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6 = arctg (x +2) + c

=∫

dxx2−4x+4+2 =

∫dx

(x−2)2+(√2)2 =

]

=∫

dtt2+(√2)2 =

1√2 arctg

t√2 + c =

1√2 arctg

x−2√2 + c, x ∈R, c∈R.

∫dx

x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c

=∫

dxx2−4x+4−2 =

∫dx

(x−2)2−(√2)2 =

[Subst. t =x−2 x ∈R−{2±

√2}

dt =dx t∈R−{±√2}

]

=∫

dtt2−(√2)2 =

12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,

x ∈R−{2±√2}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6 = arctg (x +2) + c

=∫

dxx2−4x+4+2 =

∫dx

(x−2)2+(√2)2 =

]

=∫

dtt2+(√2)2 =

1√2 arctg

t√2 + c =

1√2 arctg

x−2√2 + c, x ∈R, c∈R.

∫dx

x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c

=∫

dxx2−4x+4−2 =

∫dx

(x−2)2−(√2)2 =

[Subst. t =x−2 x ∈R−{2±

√2}

dt =dx t∈R−{±√2}

]

=∫

dtt2−(√2)2 =

12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,

x ∈R−{2±√2}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6 = arctg (x +2) + c

=∫

dxx2−4x+4+2 =

∫dx

(x−2)2+(√2)2 =

]

=∫

dtt2+(√2)2 =

1√2 arctg

t√2 + c =

1√2 arctg

x−2√2 + c, x ∈R, c∈R.

∫dx

x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c

=∫

dxx2−4x+4−2 =

∫dx

(x−2)2−(√2)2 =

[Subst. t =x−2 x ∈R−{2±

√2}

dt =dx t∈R−{±√2}

]

=∫

dtt2−(√2)2 =

12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,

x ∈R−{2±√2}, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+4 = −1

x+2 + c

=∫

dx(x+2)2 =

[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}

]

=∫

dtt2 =

∫t−2 dt = t

−2+1

−2+1 +c =−1t +c = −

1x+2 +c, x ∈R−{−2}, c∈R.∫

dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c

=∫

dx√(x+2)2

=∫

dx|x+2|=

]=∫

dt|t|

t>0 =∫

dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,

t

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+4 = −1

x+2 + c

=∫

dx(x+2)2 =

]

=∫

dtt2 =

∫t−2 dt = t

−2+1

−2+1 +c =−1t +c = −

1x+2 +c, x ∈R−{−2}, c∈R.∫

dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c

=∫

dx√(x+2)2

=∫

dx|x+2|=

]=∫

dt|t|

t>0 =∫

dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,

t

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+4 = −1

x+2 + c

=∫

dx(x+2)2 =

]

=∫

dtt2 =

∫t−2 dt = t

−2+1

−2+1 +c =−1t +c = −

1x+2 +c, x ∈R−{−2}, c∈R.∫

dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c

=∫

dx√(x+2)2

=∫

dx|x+2|=

]=∫

dt|t|

t>0 =∫

dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,

t

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+4 = −1

x+2 + c

=∫

dx(x+2)2 =

]

=∫

dtt2 =

∫t−2 dt = t

−2+1

−2+1 +c =−1t +c = −

1x+2 +c, x ∈R−{−2}, c∈R.∫

dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c

=∫

dx√(x+2)2

=∫

dx|x+2|=

]=∫

dt|t|

t>0 =∫

dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,

t

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+4 = −1

x+2 + c

=∫

dx(x+2)2 =

]

=∫

dtt2 =

∫t−2 dt = t

−2+1

−2+1 +c =−1t +c = −

1x+2 +c, x ∈R−{−2}, c∈R.∫

dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c

=∫

dx√(x+2)2

=∫

dx|x+2|=

]=∫

dt|t|

t>0 =∫

dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,

t

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+4 = −1

x+2 + c

=∫

dx(x+2)2 =

]

=∫

dtt2 =

∫t−2 dt = t

−2+1

−2+1 +c =−1t +c = −

1x+2 +c, x ∈R−{−2}, c∈R.∫

dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c

=∫

dx√(x+2)2

=∫

dx|x+2|=

]=∫

dt|t|

t>0 =∫

dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,

t

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+4 = −1

x+2 + c

=∫

dx(x+2)2 =

]

=∫

dtt2 =

∫t−2 dt = t

−2+1

−2+1 +c =−1t +c = −

1x+2 +c, x ∈R−{−2}, c∈R.∫

dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c

=∫

dx√(x+2)2

=∫

dx|x+2|=

]=∫

dt|t|

t>0 =∫

dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,

t

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

Riešené príklady – 107, 108∫dx√

x2+4x+5 = ln (x +2+√

x2+4x +5)+c

=∫

dx√x2+4x+4+1 =

∫dx√

(x+2)2+1=[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

dt√t2+1 = ln (t+

√t2+1)+c = ln (x +2+

√(x +2)2+1)+c

= ln (x +2+√

x2+4x +5)+c, x ∈R, c∈R.∫dx√

x2+4x+3 = ln |x +2+√

x2+4x +3|+c

=∫

dx√x2+4x+4−1

=∫

dx√(x+2)2−1

=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx x ∈(−1;∞) , t∈(1;∞)

]

=∫

dt√t2−1

= ln |t+√

t2−1|+c = ln |x +2+√

(x +2)2−1|+c

= ln |x +2+√

x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = ln (x +2+√

x2+4x +5)+c

=∫

dx√x2+4x+4+1 =

∫dx√

(x+2)2+1=[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

dt√t2+1 = ln (t+

√t2+1)+c = ln (x +2+

√(x +2)2+1)+c

= ln (x +2+√

x2+4x +5)+c, x ∈R, c∈R.∫dx√

x2+4x+3 = ln |x +2+√

x2+4x +3|+c

=∫

dx√x2+4x+4−1

=∫

dx√(x+2)2−1

=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx x ∈(−1;∞) , t∈(1;∞)

]

=∫

dt√t2−1

= ln |t+√

t2−1|+c = ln |x +2+√

(x +2)2−1|+c

= ln |x +2+√

x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = ln (x +2+√

x2+4x +5)+c

=∫

dx√x2+4x+4+1 =

∫dx√

(x+2)2+1=[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

dt√t2+1 = ln (t+

√t2+1)+c = ln (x +2+

√(x +2)2+1)+c

= ln (x +2+√

x2+4x +5)+c, x ∈R, c∈R.∫dx√

x2+4x+3 = ln |x +2+√

x2+4x +3|+c

=∫

dx√x2+4x+4−1

=∫

dx√(x+2)2−1

=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx x ∈(−1;∞) , t∈(1;∞)

]

=∫

dt√t2−1

= ln |t+√

t2−1|+c = ln |x +2+√

(x +2)2−1|+c

= ln |x +2+√

x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = ln (x +2+√

x2+4x +5)+c

=∫

dx√x2+4x+4+1 =

∫dx√

(x+2)2+1=[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

dt√t2+1 = ln (t+

√t2+1)+c = ln (x +2+

√(x +2)2+1)+c

= ln (x +2+√

x2+4x +5)+c, x ∈R, c∈R.∫dx√

x2+4x+3 = ln |x +2+√

x2+4x +3|+c

=∫

dx√x2+4x+4−1

=∫

dx√(x+2)2−1

=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx x ∈(−1;∞) , t∈(1;∞)

]

=∫

dt√t2−1

= ln |t+√

t2−1|+c = ln |x +2+√

(x +2)2−1|+c

= ln |x +2+√

x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = ln (x +2+√

x2+4x +5)+c

=∫

dx√x2+4x+4+1 =

∫dx√

(x+2)2+1=[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

dt√t2+1 = ln (t+

√t2+1)+c = ln (x +2+

√(x +2)2+1)+c

= ln (x +2+√

x2+4x +5)+c, x ∈R, c∈R.∫dx√

x2+4x+3 = ln |x +2+√

x2+4x +3|+c

=∫

dx√x2+4x+4−1

=∫

dx√(x+2)2−1

=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx x ∈(−1;∞) , t∈(1;∞)

]

=∫

dt√t2−1

= ln |t+√

t2−1|+c = ln |x +2+√

(x +2)2−1|+c

= ln |x +2+√

x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = ln (x +2+√

x2+4x +5)+c

=∫

dx√x2+4x+4+1 =

∫dx√

(x+2)2+1=[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

dt√t2+1 = ln (t+

√t2+1)+c = ln (x +2+

√(x +2)2+1)+c

= ln (x +2+√

x2+4x +5)+c, x ∈R, c∈R.∫dx√

x2+4x+3 = ln |x +2+√

x2+4x +3|+c

=∫

dx√x2+4x+4−1

=∫

dx√(x+2)2−1

=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx x ∈(−1;∞) , t∈(1;∞)

]

=∫

dt√t2−1

= ln |t+√

t2−1|+c = ln |x +2+√

(x +2)2−1|+c

= ln |x +2+√

x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2+4x+5 = ln (x +2+√

x2+4x +5)+c

=∫

dx√x2+4x+4+1 =

∫dx√

(x+2)2+1=[Subst. t =x +2 x ∈R

dt =dx t∈R

]

=∫

dt√t2+1 = ln (t+

√t2+1)+c = ln (x +2+

√(x +2)2+1)+c

= ln (x +2+√

x2+4x +5)+c, x ∈R, c∈R.∫dx√

x2+4x+3 = ln |x +2+√

x2+4x +3|+c

=∫

dx√x2+4x+4−1

=∫

dx√(x+2)2−1

=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx x ∈(−1;∞) , t∈(1;∞)

]

=∫

dt√t2−1

= ln |t+√

t2−1|+c = ln |x +2+√

(x +2)2−1|+c

= ln |x +2+√

x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6= ln (x−2+

√x2−4x +6)+c

=∫

dx√x2−4x+4+2

=∫

dx√(x−2)2+(

√2)2

=[Subst. t =x−2 x ∈R

dt =dx t∈R

]

=∫

dt√t2+(√2)2

= ln(t+√

t2+(√2)2)+c = ln (x +2+

√(x−2)2+4)+c

= ln (x−2+√

x2−4x +6)+c, x ∈R, c∈R.∫dx√

x2−4x+2= ln |x−2+

√x2−4x +2|+c

=∫

dx√x2−4x+4−2

=∫

dx√(x−2)2−(

√2)2

=[Subst. t =x−2 x ∈(−∞; 2−

√2), t∈(−∞;−

√2)

dt =dx x ∈(2+√2;∞), t∈(

√2;∞)

]

=∫

dt√t2−(√2)2

= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c

= ln |x−2+√

x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+

√2;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6= ln (x−2+

√x2−4x +6)+c

=∫

dx√x2−4x+4+2

=∫

dx√(x−2)2+(

√2)2

dt =dx t∈R

]

=∫

dt√t2+(√2)2

= ln(t+√

t2+(√2)2)+c = ln (x +2+

√(x−2)2+4)+c

= ln (x−2+√

x2−4x +6)+c, x ∈R, c∈R.∫dx√

x2−4x+2= ln |x−2+

√x2−4x +2|+c

=∫

dx√x2−4x+4−2

=∫

dx√(x−2)2−(

√2)2

=[Subst. t =x−2 x ∈(−∞; 2−

√2), t∈(−∞;−

√2)

dt =dx x ∈(2+√2;∞), t∈(

√2;∞)

]

=∫

dt√t2−(√2)2

= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c

= ln |x−2+√

x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+

√2;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6= ln (x−2+

√x2−4x +6)+c

=∫

dx√x2−4x+4+2

=∫

dx√(x−2)2+(

√2)2

dt =dx t∈R

]

=∫

dt√t2+(√2)2

= ln(t+√

t2+(√2)2)+c = ln (x +2+

√(x−2)2+4)+c

= ln (x−2+√

x2−4x +6)+c, x ∈R, c∈R.∫dx√

x2−4x+2= ln |x−2+

√x2−4x +2|+c

=∫

dx√x2−4x+4−2

=∫

dx√(x−2)2−(

√2)2

=[Subst. t =x−2 x ∈(−∞; 2−

√2), t∈(−∞;−

√2)

dt =dx x ∈(2+√2;∞), t∈(

√2;∞)

]

=∫

dt√t2−(√2)2

= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c

= ln |x−2+√

x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+

√2;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6= ln (x−2+

√x2−4x +6)+c

=∫

dx√x2−4x+4+2

=∫

dx√(x−2)2+(

√2)2

dt =dx t∈R

]

=∫

dt√t2+(√2)2

= ln(t+√

t2+(√2)2)+c = ln (x +2+

√(x−2)2+4)+c

= ln (x−2+√

x2−4x +6)+c, x ∈R, c∈R.∫dx√

x2−4x+2= ln |x−2+

√x2−4x +2|+c

=∫

dx√x2−4x+4−2

=∫

dx√(x−2)2−(

√2)2

=[Subst. t =x−2 x ∈(−∞; 2−

√2), t∈(−∞;−

√2)

dt =dx x ∈(2+√2;∞), t∈(

√2;∞)

]

=∫

dt√t2−(√2)2

= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c

= ln |x−2+√

x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+

√2;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6= ln (x−2+

√x2−4x +6)+c

=∫

dx√x2−4x+4+2

=∫

dx√(x−2)2+(

√2)2

dt =dx t∈R

]

=∫

dt√t2+(√2)2

= ln(t+√

t2+(√2)2)+c = ln (x +2+

√(x−2)2+4)+c

= ln (x−2+√

x2−4x +6)+c, x ∈R, c∈R.∫dx√

x2−4x+2= ln |x−2+

√x2−4x +2|+c

=∫

dx√x2−4x+4−2

=∫

dx√(x−2)2−(

√2)2

=[Subst. t =x−2 x ∈(−∞; 2−

√2), t∈(−∞;−

√2)

dt =dx x ∈(2+√2;∞), t∈(

√2;∞)

]

=∫

dt√t2−(√2)2

= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c

= ln |x−2+√

x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+

√2;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6= ln (x−2+

√x2−4x +6)+c

=∫

dx√x2−4x+4+2

=∫

dx√(x−2)2+(

√2)2

dt =dx t∈R

]

=∫

dt√t2+(√2)2

= ln(t+√

t2+(√2)2)+c = ln (x +2+

√(x−2)2+4)+c

= ln (x−2+√

x2−4x +6)+c, x ∈R, c∈R.∫dx√

x2−4x+2= ln |x−2+

√x2−4x +2|+c

=∫

dx√x2−4x+4−2

=∫

dx√(x−2)2−(

√2)2

=[Subst. t =x−2 x ∈(−∞; 2−

√2), t∈(−∞;−

√2)

dt =dx x ∈(2+√2;∞), t∈(

√2;∞)

]

=∫

dt√t2−(√2)2

= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c

= ln |x−2+√

x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+

√2;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10

x2−4x+6= ln (x−2+

√x2−4x +6)+c

=∫

dx√x2−4x+4+2

=∫

dx√(x−2)2+(

√2)2

dt =dx t∈R

]

=∫

dt√t2+(√2)2

= ln(t+√

t2+(√2)2)+c = ln (x +2+

√(x−2)2+4)+c

= ln (x−2+√

x2−4x +6)+c, x ∈R, c∈R.∫dx√

x2−4x+2= ln |x−2+

√x2−4x +2|+c

=∫

dx√x2−4x+4−2

=∫

dx√(x−2)2−(

√2)2

=[Subst. t =x−2 x ∈(−∞; 2−

√2), t∈(−∞;−

√2)

dt =dx x ∈(2+√2;∞), t∈(

√2;∞)

]

=∫

dt√t2−(√2)2

= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c

= ln |x−2+√

x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+

√2;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

−x2+4x−5nemá riešenie

=∫

dx√−(x2−4x+5)

=∫

dx√−(x2−4x+4+1)

=∫

dx√−(x−2)2−1

=[−(x−2)2−1≥1>0

pre všetky x ∈R

].

Neexistuje riešenie pre žiadne x ∈R.∫dx√

−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2

=∫

dx√−(x2−4x+3)

=∫

dx√−(x2−4x+4−1)

=∫

dx√−(x−2)2+1

=∫

dx√1−(x−2)2

=[ Subst. t =x−2 x ∈(1; 3)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

=∫

dx√−(x2−4x+5)

=∫

dx√−(x2−4x+4+1)

=∫

dx√−(x−2)2−1

=[−(x−2)2−1≥1>0

pre všetky x ∈R

].

=∫

dx√−(x2−4x+3)

=∫

dx√−(x2−4x+4−1)

=∫

dx√−(x−2)2+1

=∫

dx√1−(x−2)2

=[ Subst. t =x−2 x ∈(1; 3)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

=∫

dx√−(x2−4x+5)

=∫

dx√−(x2−4x+4+1)

=∫

dx√−(x−2)2−1

=[−(x−2)2−1≥1>0

pre všetky x ∈R

].

=∫

dx√−(x2−4x+3)

=∫

dx√−(x2−4x+4−1)

=∫

dx√−(x−2)2+1

=∫

dx√1−(x−2)2

=[ Subst. t =x−2 x ∈(1; 3)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

=∫

dx√−(x2−4x+5)

=∫

dx√−(x2−4x+4+1)

=∫

dx√−(x−2)2−1

=[−(x−2)2−1≥1>0

pre všetky x ∈R

].

=∫

dx√−(x2−4x+3)

=∫

dx√−(x2−4x+4−1)

=∫

dx√−(x−2)2+1

=∫

dx√1−(x−2)2

=[ Subst. t =x−2 x ∈(1; 3)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

=∫

dx√−(x2−4x+5)

=∫

dx√−(x2−4x+4+1)

=∫

dx√−(x−2)2−1

=[−(x−2)2−1≥1>0

pre všetky x ∈R

].

=∫

dx√−(x2−4x+3)

=∫

dx√−(x2−4x+4−1)

=∫

dx√−(x−2)2+1

=∫

dx√1−(x−2)2

=[ Subst. t =x−2 x ∈(1; 3)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

=∫

dx√−(x2−4x+5)

=∫

dx√−(x2−4x+4+1)

=∫

dx√−(x−2)2−1

=[−(x−2)2−1≥1>0

pre všetky x ∈R

].

=∫

dx√−(x2−4x+3)

=∫

dx√−(x2−4x+4−1)

=∫

dx√−(x−2)2+1

=∫

dx√1−(x−2)2

=[ Subst. t =x−2 x ∈(1; 3)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

=∫

dx√−(x2−4x+5)

=∫

dx√−(x2−4x+4+1)

=∫

dx√−(x−2)2−1

=[−(x−2)2−1≥1>0

pre všetky x ∈R

].

=∫

dx√−(x2−4x+3)

=∫

dx√−(x2−4x+4−1)

=∫

dx√−(x−2)2+1

=∫

dx√1−(x−2)2

=[ Subst. t =x−2 x ∈(1; 3)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

=∫

dx√−(x2−4x+5)

=∫

dx√−(x2−4x+4+1)

=∫

dx√−(x−2)2−1

=[−(x−2)2−1≥1>0

pre všetky x ∈R

].

=∫

dx√−(x2−4x+3)

=∫

dx√−(x2−4x+4−1)

=∫

dx√−(x−2)2+1

=∫

dx√1−(x−2)2

=[ Subst. t =x−2 x ∈(1; 3)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

= arcsin (x−1)+c1 = − arccos (x−1)+c2

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

x(x−1)=∫

dx√x2−x

= ln |x− 12 +√

x2−x |+c1 = ln |2x−1+√

x2−x |+c2

=∫

dx√x2−2 x2 +

14−

14

=∫

dx√(x− 12 )2−(

12 )2

=[Subst. t =x− 12 x ∈(−∞; 0) , t∈

(−∞;− 12

)dt =dx x ∈(1;∞) , t∈

( 12 ;∞

) ]

=∫

dt√t2−( 12 )2

= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1

= ln |2x−1+√

x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√

x(2−x)=∫

dx√−x2+2x

=∫

dx√−(x2−2x+1)+1

=∫

dx√1−(x−1)2

=[ Subst. t =x−1 x ∈(0; 2)

dt =dx t∈(−1; 1)

]=∫

dt√1−t2

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

Riešené príklady – 115, 116∫ √x2+4x +5 dx = (x+2)

√x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c

=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R

dt =dx =(x +2)2+1= t2+1 t∈R

]=∫ √

t2+1 dt = t√

t2+12 +

12

∫dt√t2+1

= t√

t2+12 +

ln (t+√

t2+1)2 +c =

(x+2)√

(x+2)2+12 +

ln (x+2+√

(x+2)2+1)2 +c

= (x+2)√

x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c, x ∈R, c∈R.∫ √

x2+4x +3 dx = (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c

=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−1 dt

= t√

t2−12 −

12

∫dt√t2−1

= t√

t2−12 −

ln |t+√

t2−1|2 +c

= (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c

=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R

dt =dx =(x +2)2+1= t2+1 t∈R

]=∫ √

t2+1 dt = t√

t2+12 +

12

∫dt√t2+1

= t√

t2+12 +

ln (t+√

t2+1)2 +c =

(x+2)√

(x+2)2+12 +

ln (x+2+√

(x+2)2+1)2 +c

= (x+2)√

x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c, x ∈R, c∈R.∫ √

x2+4x +3 dx = (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c

=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−1 dt

= t√

t2−12 −

12

∫dt√t2−1

= t√

t2−12 −

ln |t+√

t2−1|2 +c

= (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c

=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R

dt =dx =(x +2)2+1= t2+1 t∈R

]=∫ √

t2+1 dt = t√

t2+12 +

12

∫dt√t2+1

= t√

t2+12 +

ln (t+√

t2+1)2 +c =

(x+2)√

(x+2)2+12 +

ln (x+2+√

(x+2)2+1)2 +c

= (x+2)√

x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c, x ∈R, c∈R.∫ √

x2+4x +3 dx = (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c

=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−1 dt

= t√

t2−12 −

12

∫dt√t2−1

= t√

t2−12 −

ln |t+√

t2−1|2 +c

= (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c

=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R

dt =dx =(x +2)2+1= t2+1 t∈R

]=∫ √

t2+1 dt = t√

t2+12 +

12

∫dt√t2+1

= t√

t2+12 +

ln (t+√

t2+1)2 +c =

(x+2)√

(x+2)2+12 +

ln (x+2+√

(x+2)2+1)2 +c

= (x+2)√

x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c, x ∈R, c∈R.∫ √

x2+4x +3 dx = (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c

=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−1 dt

= t√

t2−12 −

12

∫dt√t2−1

= t√

t2−12 −

ln |t+√

t2−1|2 +c

= (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c

=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R

dt =dx =(x +2)2+1= t2+1 t∈R

]=∫ √

t2+1 dt = t√

t2+12 +

12

∫dt√t2+1

= t√

t2+12 +

ln (t+√

t2+1)2 +c =

(x+2)√

(x+2)2+12 +

ln (x+2+√

(x+2)2+1)2 +c

= (x+2)√

x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c, x ∈R, c∈R.∫ √

x2+4x +3 dx = (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c

=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−1 dt

= t√

t2−12 −

12

∫dt√t2−1

= t√

t2−12 −

ln |t+√

t2−1|2 +c

= (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c

=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R

dt =dx =(x +2)2+1= t2+1 t∈R

]=∫ √

t2+1 dt = t√

t2+12 +

12

∫dt√t2+1

= t√

t2+12 +

ln (t+√

t2+1)2 +c =

(x+2)√

(x+2)2+12 +

ln (x+2+√

(x+2)2+1)2 +c

= (x+2)√

x2+4x+52 +

ln (x+2+√

x2+4x+5)2 +c, x ∈R, c∈R.∫ √

x2+4x +3 dx = (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c

=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−1 dt

= t√

t2−12 −

12

∫dt√t2−1

= t√

t2−12 −

ln |t+√

t2−1|2 +c

= (x+2)√

x2+4x+32 −

ln |x+2+√

x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

Riešené príklady – 117, 118∫ √x2−4x +6 dx = (x−2)

√x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c

=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R

dt =dx =(x−2)2+2= t2+(√2)2 t∈R

]=∫√

t2+(√2)2 dt

= t√

t2+(√2)2

2 +12

∫dt√

t2+(√2)2

= t√

t2+(√2)2

2 +ln (t+√

t2+(√2)2)

2 +c

= (x−2)√

x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c, x ∈R, c∈R.∫ √

x2−4x +2 dx = (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c

=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−(√2)2 dt

= t√

t2−(√2)2

2 −12

∫dt√

t2−(√2)2

= t√

t2−(√2)2

2 −ln∣∣t+√t2−(√2)2∣∣

2 +c

= (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c, x ∈R−

〈2−√2; 2+

√2〉, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c

=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R

dt =dx =(x−2)2+2= t2+(√2)2 t∈R

]=∫√

t2+(√2)2 dt

= t√

t2+(√2)2

2 +12

∫dt√

t2+(√2)2

= t√

t2+(√2)2

2 +ln (t+√

t2+(√2)2)

2 +c

= (x−2)√

x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c, x ∈R, c∈R.∫ √

x2−4x +2 dx = (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c

=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−(√2)2 dt

= t√

t2−(√2)2

2 −12

∫dt√

t2−(√2)2

= t√

t2−(√2)2

2 −ln∣∣t+√t2−(√2)2∣∣

2 +c

= (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c, x ∈R−

〈2−√2; 2+

√2〉, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c

=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R

dt =dx =(x−2)2+2= t2+(√2)2 t∈R

]=∫√

t2+(√2)2 dt

= t√

t2+(√2)2

2 +12

∫dt√

t2+(√2)2

= t√

t2+(√2)2

2 +ln (t+√

t2+(√2)2)

2 +c

= (x−2)√

x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c, x ∈R, c∈R.∫ √

x2−4x +2 dx = (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c

=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−(√2)2 dt

= t√

t2−(√2)2

2 −12

∫dt√

t2−(√2)2

= t√

t2−(√2)2

2 −ln∣∣t+√t2−(√2)2∣∣

2 +c

= (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c, x ∈R−

〈2−√2; 2+

√2〉, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c

=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R

dt =dx =(x−2)2+2= t2+(√2)2 t∈R

]=∫√

t2+(√2)2 dt

= t√

t2+(√2)2

2 +12

∫dt√

t2+(√2)2

= t√

t2+(√2)2

2 +ln (t+√

t2+(√2)2)

2 +c

= (x−2)√

x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c, x ∈R, c∈R.∫ √

x2−4x +2 dx = (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c

=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−(√2)2 dt

= t√

t2−(√2)2

2 −12

∫dt√

t2−(√2)2

= t√

t2−(√2)2

2 −ln∣∣t+√t2−(√2)2∣∣

2 +c

= (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c, x ∈R−

〈2−√2; 2+

√2〉, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c

=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R

dt =dx =(x−2)2+2= t2+(√2)2 t∈R

]=∫√

t2+(√2)2 dt

= t√

t2+(√2)2

2 +12

∫dt√

t2+(√2)2

= t√

t2+(√2)2

2 +ln (t+√

t2+(√2)2)

2 +c

= (x−2)√

x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c, x ∈R, c∈R.∫ √

x2−4x +2 dx = (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c

=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−(√2)2 dt

= t√

t2−(√2)2

2 −12

∫dt√

t2−(√2)2

= t√

t2−(√2)2

2 −ln∣∣t+√t2−(√2)2∣∣

2 +c

= (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c, x ∈R−

〈2−√2; 2+

√2〉, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c

=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R

dt =dx =(x−2)2+2= t2+(√2)2 t∈R

]=∫√

t2+(√2)2 dt

= t√

t2+(√2)2

2 +12

∫dt√

t2+(√2)2

= t√

t2+(√2)2

2 +ln (t+√

t2+(√2)2)

2 +c

= (x−2)√

x2−4x+62 +

ln (x−2+√

x2−4x+6)2 +c, x ∈R, c∈R.∫ √

x2−4x +2 dx = (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c

=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)

dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)

]=∫ √

t2−(√2)2 dt

= t√

t2−(√2)2

2 −12

∫dt√

t2−(√2)2

= t√

t2−(√2)2

2 −ln∣∣t+√t2−(√2)2∣∣

2 +c

= (x−2)√

x2+4x+22 −

ln |x−2+√

x2−4x+2|2 +c, x ∈R−

〈2−√2; 2+

√2〉, c∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

Riešené príklady – 119, 120∫ √−x2+4x−3 dx = (x−2)

√−x2+4x−32 +

arcsin (x−2)2 +c1

=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)

dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt

= t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2

= (x−2)√−x2+4x−32 +

arcsin (x−2)2 +c1=

(x−2)√−x2+4x−32 −

arccos (x−2)2 +c2,

x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =

∫ √−x2+2x dx = (x−2)

√−x2+2x2 +

arcsin (x−2)2 +c1

=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)

dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt = t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2=

(x−2)√−x2+2x2 +

arcsin (x−2)2 +c1

= (x−2)√−x2+2x2 −

arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√−x2+4x−32 +

arcsin (x−2)2 +c1

=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)

dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt

= t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2

= (x−2)√−x2+4x−32 +

(x−2)√−x2+4x−32 −

x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =

∫ √−x2+2x dx = (x−2)

√−x2+2x2 +

arcsin (x−2)2 +c1

=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)

dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt = t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2=

(x−2)√−x2+2x2 +

arcsin (x−2)2 +c1

= (x−2)√−x2+2x2 −

arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√−x2+4x−32 +

arcsin (x−2)2 +c1

=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)

dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt

= t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2

= (x−2)√−x2+4x−32 +

(x−2)√−x2+4x−32 −

x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =

∫ √−x2+2x dx = (x−2)

√−x2+2x2 +

arcsin (x−2)2 +c1

=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)

dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt = t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2=

(x−2)√−x2+2x2 +

arcsin (x−2)2 +c1

= (x−2)√−x2+2x2 −

arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√−x2+4x−32 +

arcsin (x−2)2 +c1

=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)

dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt

= t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2

= (x−2)√−x2+4x−32 +

(x−2)√−x2+4x−32 −

x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =

∫ √−x2+2x dx = (x−2)

√−x2+2x2 +

arcsin (x−2)2 +c1

=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)

dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt = t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2=

(x−2)√−x2+2x2 +

arcsin (x−2)2 +c1

= (x−2)√−x2+2x2 −

arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√−x2+4x−32 +

arcsin (x−2)2 +c1

=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)

dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt

= t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2

= (x−2)√−x2+4x−32 +

(x−2)√−x2+4x−32 −

x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =

∫ √−x2+2x dx = (x−2)

√−x2+2x2 +

arcsin (x−2)2 +c1

=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)

dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt = t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2=

(x−2)√−x2+2x2 +

arcsin (x−2)2 +c1

= (x−2)√−x2+2x2 −

arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20

√−x2+4x−32 +

arcsin (x−2)2 +c1

=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)

dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt

= t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2

= (x−2)√−x2+4x−32 +

(x−2)√−x2+4x−32 −

x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =

∫ √−x2+2x dx = (x−2)

√−x2+2x2 +

arcsin (x−2)2 +c1

=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)

dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)

]=∫ √

1−t2 dt = t√

1−t22 +

12

∫dt√1−t2

= t√

1−t22 +

arcsin t2 +c1=

t√

1−t22 −

arccos t2 +c2=

(x−2)√−x2+2x2 +

arcsin (x−2)2 +c1

= (x−2)√−x2+2x2 −

arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30

Riešené príklady – 121

In =∫

dx(x2+a2)n =

2n−32a2(n−1) In−1+

x2a2(n−1)(x2+a2)n−1 , I1=

1a arctg

xa +c, a>0, n∈N

= 1a2∫

a2 dx(x2+a2)n =

1a2

∫(x2+a2−x2) dx

(x2+a2)n =1a2

∫(x2+a2) dx(x2+a2)n −

1a2

∫x2 dx

(x2+a2)n

= 1a2∫

dx(x2+a2)n−1 −

12a2

∫x ·2x dx

(x2+a2)n =[

u = x u′= 1v ′= 2x(x2+a2)n v =

(x2+a2)1−n1−n =

1(1−n)(x2+a2)n−1

]

= 1a2 In−1 −12a2

[x

(1−n)(x2+a2)n−1 −1

1−n

∫dx

(x2+a2)n−1

]= 1a2 In−1 −

x2a2(1−n)(x2+a2)n−1 +

12a2(1−n) In−1

= 12a2 (2+1

1−n )In−1 +x

2a2(n−1)(x2+a2)n−1 =3−2n

2a2(1−n) In−1 +x

2a2(n−1)(x2+a2)n−1

= 2n−32a2(n−1) In−1 +x

2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .

I1=∫

dxx2+a2 =

1a arctg

xa + c, x ∈R, c∈R, n=1.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30

In =∫

dx(x2+a2)n =

2n−32a2(n−1) In−1+

x2a2(n−1)(x2+a2)n−1 , I1=

1a arctg

xa +c, a>0, n∈N

= 1a2∫

a2 dx(x2+a2)n =

1a2

∫(x2+a2−x2) dx

(x2+a2)n =1a2

∫(x2+a2) dx(x2+a2)n −

1a2

∫x2 dx

(x2+a2)n

= 1a2∫

dx(x2+a2)n−1 −

12a2

∫x ·2x dx

(x2+a2)n =[

u = x u′= 1v ′= 2x(x2+a2)n v =

(x2+a2)1−n1−n =

1(1−n)(x2+a2)n−1

]

= 1a2 In−1 −12a2

[x

(1−n)(x2+a2)n−1 −1

1−n

∫dx

(x2+a2)n−1

]= 1a2 In−1 −

x2a2(1−n)(x2+a2)n−1 +

12a2(1−n) In−1

= 12a2 (2+1

1−n )In−1 +x

2a2(n−1)(x2+a2)n−1 =3−2n

2a2(1−n) In−1 +x

2a2(n−1)(x2+a2)n−1

= 2n−32a2(n−1) In−1 +x

2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .

I1=∫

dxx2+a2 =

1a arctg

xa + c, x ∈R, c∈R, n=1.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30

In =∫

dx(x2+a2)n =

2n−32a2(n−1) In−1+

x2a2(n−1)(x2+a2)n−1 , I1=

1a arctg

xa +c, a>0, n∈N

= 1a2∫

a2 dx(x2+a2)n =

1a2

∫(x2+a2−x2) dx

(x2+a2)n =1a2

∫(x2+a2) dx(x2+a2)n −

1a2

∫x2 dx

(x2+a2)n

= 1a2∫

dx(x2+a2)n−1 −

12a2

∫x ·2x dx

(x2+a2)n =[

u = x u′= 1v ′= 2x(x2+a2)n v =

(x2+a2)1−n1−n =

1(1−n)(x2+a2)n−1

]

= 1a2 In−1 −12a2

[x

(1−n)(x2+a2)n−1 −1

1−n

∫dx

(x2+a2)n−1

]= 1a2 In−1 −

x2a2(1−n)(x2+a2)n−1 +

12a2(1−n) In−1

= 12a2 (2+1

1−n )In−1 +x

2a2(n−1)(x2+a2)n−1 =3−2n

2a2(1−n) In−1 +x

2a2(n−1)(x2+a2)n−1

= 2n−32a2(n−1) In−1 +x

2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .

I1=∫

dxx2+a2 =

1a arctg

xa + c, x ∈R, c∈R, n=1.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30

In =∫

dx(x2+a2)n =

2n−32a2(n−1) In−1+

x2a2(n−1)(x2+a2)n−1 , I1=

1a arctg

xa +c, a>0, n∈N

= 1a2∫

a2 dx(x2+a2)n =

1a2

∫(x2+a2−x2) dx

(x2+a2)n =1a2

∫(x2+a2) dx(x2+a2)n −

1a2

∫x2 dx

(x2+a2)n

= 1a2∫

dx(x2+a2)n−1 −

12a2

∫x ·2x dx

(x2+a2)n =[

u = x u′= 1v ′= 2x(x2+a2)n v =

(x2+a2)1−n1−n =

1(1−n)(x2+a2)n−1

]

= 1a2 In−1 −12a2

[x

(1−n)(x2+a2)n−1 −1

1−n

∫dx

(x2+a2)n−1

]= 1a2 In−1 −

x2a2(1−n)(x2+a2)n−1 +

12a2(1−n) In−1

= 12a2 (2+1

1−n )In−1 +x

2a2(n−1)(x2+a2)n−1 =3−2n

2a2(1−n) In−1 +x

2a2(n−1)(x2+a2)n−1

= 2n−32a2(n−1) In−1 +x

2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .

I1=∫

dxx2+a2 =

1a arctg

xa + c, x ∈R, c∈R, n=1.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30

In =∫

dx(x2+a2)n =

2n−32a2(n−1) In−1+

x2a2(n−1)(x2+a2)n−1 , I1=

1a arctg

xa +c, a>0, n∈N

= 1a2∫

a2 dx(x2+a2)n =

1a2

∫(x2+a2−x2) dx

(x2+a2)n =1a2

∫(x2+a2) dx(x2+a2)n −

1a2

∫x2 dx

(x2+a2)n

= 1a2∫

dx(x2+a2)n−1 −

12a2

∫x ·2x dx

(x2+a2)n =[

u = x u′= 1v ′= 2x(x2+a2)n v =

(x2+a2)1−n1−n =

1(1−n)(x2+a2)n−1

]

= 1a2 In−1 −12a2

[x

(1−n)(x2+a2)n−1 −1

1−n

∫dx

(x2+a2)n−1

]= 1a2 In−1 −

x2a2(1−n)(x2+a2)n−1 +

12a2(1−n) In−1

= 12a2 (2+1

1−n )In−1 +x

2a2(n−1)(x2+a2)n−1 =3−2n

2a2(1−n) In−1 +x

2a2(n−1)(x2+a2)n−1

= 2n−32a2(n−1) In−1 +x

2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .

I1=∫

dxx2+a2 =

1a arctg

xa + c, x ∈R, c∈R, n=1.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30

In =∫

dx(x2+a2)n =

2n−32a2(n−1) In−1+

x2a2(n−1)(x2+a2)n−1 , I1=

1a arctg

xa +c, a>0, n∈N

= 1a2∫

a2 dx(x2+a2)n =

1a2

∫(x2+a2−x2) dx

(x2+a2)n =1a2

∫(x2+a2) dx(x2+a2)n −

1a2

∫x2 dx

(x2+a2)n

= 1a2∫

dx(x2+a2)n−1 −

12a2

∫x ·2x dx

(x2+a2)n =[

u = x u′= 1v ′= 2x(x2+a2)n v =

(x2+a2)1−n1−n =

1(1−n)(x2+a2)n−1

]

= 1a2 In−1 −12a2

[x

(1−n)(x2+a2)n−1 −1

1−n

∫dx

(x2+a2)n−1

]= 1a2 In−1 −

x2a2(1−n)(x2+a2)n−1 +

12a2(1−n) In−1

= 12a2 (2+1

1−n )In−1 +x

2a2(n−1)(x2+a2)n−1 =3−2n

2a2(1−n) In−1 +x

2a2(n−1)(x2+a2)n−1

= 2n−32a2(n−1) In−1 +x

2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .

I1=∫

dxx2+a2 =

1a arctg

xa + c, x ∈R, c∈R, n=1.

Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30

In =∫

dx(x2+a2)n =

2n−32a2(n−1) In−1+

x2a2(n−1)(x2+a2)n−1 , I1=

1a arctg

xa +c, a>0, n∈N

= 1a2∫

a2 dx(x2+a2)n =

1a2

∫(x2+a2−x2) dx

(x2+a2)n =1a2

∫(x2+a2) dx(x2+a2)n −

1a2

∫x2 dx

(x2+a2)n

= 1a2∫

dx(x2+a2)n−1 −

12a2

∫x ·2x dx

(x2+a2)n =[

u = x u′= 1v ′= 2x(x2+a2)n v =

(x2+a2)1−n1−n =

1(1−n)(x2+a2)n−1

]

= 1a2 In−1 −12a2

[x

(1−n)(x2+a2)n−1 −1

1−n

∫dx

(x2+a2)n−1

]= 1a2 In−1 −

x2a2(1−n)(x2+a2)n−1 +

12a2(1−n) In−1

= 12a2 (2+1

1−n )In−1 +x

2a2(n−1)(x2+a2)n−1 =3−2n

2a2(1−n)

university of Žilina · obsahzoznam01112131415161718191 zoznamintegrálov–príklady101–200...

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