taller #2 integral parte 1 seguimiento 2
TRANSCRIPT
5. ∫𝑑𝑢
√𝑢2 +𝑎2 =
Sea u= atanϴ
du = a𝑠𝑒𝑐2ϴ
𝑢2+𝑎2 = 𝑎2 𝑡𝑎𝑛2ϴ+𝑎2
=𝑎2 (𝑡𝑎𝑛2ϴ+1)
=𝑎2 𝑠𝑒𝑐2ϴ
√𝑢2 + 𝑎2 = √𝑎2 + 𝑠𝑒𝑐2 𝛳
=asecϴ
∫𝑎𝑠𝑒𝑐2 𝛳𝑑𝛳
𝑎𝑠𝑒𝑐𝛳 = ∫ 𝑠𝑒𝑐𝛳𝑑𝛳 = In |𝑠𝑒𝑐𝛳 + 𝑡𝑎𝑛𝛳|
= In|√𝑢2 +𝑎2
𝑎| +
𝑢
𝑎
=In|𝑎√𝑢2 +𝑎2+ 𝑎𝑢
𝑎2 |
=In |√𝑢2 +𝑎2 +𝑢
𝑎|
=In|√𝑢2 + 𝑎2 + 𝑢| - In|𝑎|+C
=In|√𝑢2 + 𝑎2 + 𝑢|+C
4. ∫𝑥2
(𝑥2+4)2 𝑑𝑥
∫𝑥2
(𝑥2+4)2 = 𝐴𝑋 +𝐵
𝑋2 +4+
𝐶𝑋 +𝐷
(𝑋2 +4)2
𝑋2= (AX+B)(𝑋2+4)+(CX+D)
𝑋2= A𝑋3+4AX+B𝑋2+4B+CX+D
𝑋2=𝑋3(𝐴)+𝑋2(𝐵)+X(4A+C)+(4B+D)
A=0
B=1
4A+C=O; 4(0)+C=0, C=0
AB+D=0;4(1)+D=0, D=-4
∫𝑥2
(𝑥2+4) 2 = 0𝑋 +1
𝑋2 +4+
(0)𝑋−4
(𝑋2+4)2
=∫𝑑𝑥
𝑥2+4 - ∫
𝑑𝑥
(𝑥2+4)2
𝑥 = 2𝑡𝑎𝑛𝜃
𝑑𝑥 = 2𝑠𝑒𝑐2 𝜃𝑑𝜃
∫2𝑠𝑒𝑐2 𝜃𝑑𝜃
4𝑡𝑎𝑛2𝜃 + 4 − 4 ∫
2𝑠𝑒𝑐2𝜃𝑑𝜃
(4𝑡𝑎𝑛2𝜃 + 4)2
2 ∫𝑠𝑒𝑐2𝜃𝑑𝜃
4(𝑡𝑎𝑛2 𝜃 + 1) − 8 ∫
𝑠𝑒𝑐2𝜃𝑑𝜃
(4(𝑡𝑎𝑛2𝜃 + 1))2
1
2∫
𝑠𝑒𝑐2 𝜃𝑑𝜃
𝑠𝑒𝑐2 𝜃−
1
2∫
𝑠𝑒𝑐2𝜃𝑑𝜃
𝑠𝑒𝑐4 𝜃
1
2∫ 𝑑𝜃 −
1
2∫
𝑑𝜃
𝑠𝑒𝑐2 𝜃
1
2∫ 𝑑𝜃 −
1
2∫
𝑑𝜃1
𝑐𝑜𝑠2𝜃
1
2∫ 𝑑𝜃 −
1
2∫ 𝑐𝑜𝑠2 𝜃𝑑𝜃
1
2∫ 𝑑𝜃 −
1
2∫
1
2(1 + 𝑐𝑜𝑠2𝜃)𝑑𝜃
1
2∫ 𝑑𝜃 −
1
4∫ 𝑑𝜃 −
1
4∫ 𝑐𝑜𝑠2𝜃
1
4∫ 𝑑𝜃 −
1
4∫ 𝑐𝑜𝑠2𝜃
1
4(𝜃 −
1
2𝑠𝑒𝑛2𝜃) =
1
4(𝑡𝑎𝑛−1 (
𝑥
2)) −
1
8𝑠𝑒𝑛2 (𝑡𝑎𝑛−1
𝑥
2) + 𝐶
5.1
4)
∫𝑥2
(𝑥2 + 42)2𝑑𝑥
𝑥 = 2. tan 𝜃
𝑑𝑥 = 2. 𝑆𝑒𝑐2𝜃 𝑑𝜃
∫(2. tan 𝜃)2
((2. tan 𝜃)2 + 4)2 2. 𝑠𝑒𝑐2𝜃 𝑑𝜃
∫4. 𝑡𝑎𝑛2𝜃
(4. 𝑡𝑎𝑛2𝜃 + 4)2 2. 𝑆𝑒𝑐2𝜃 𝑑𝜃
8 ∫𝑡𝑎𝑛2𝜃. 𝑠𝑒𝑐2𝜃 𝑑𝜃
(4 (𝑡𝑎𝑛2𝜃 + 1))2
8 ∫𝑡𝑎𝑛2𝜃 . 𝑠𝑒𝑐2𝜃 . 𝑑𝜃
(4 . 𝑠𝑒𝑐2𝜃)2
1
2∫
𝑡𝑎𝑛2𝜃 . 𝑠𝑒𝑐2𝜃 𝑑𝜃
𝑠𝑒𝑐4𝜃
1
2∫
𝑡𝑎𝑛2𝜃 𝑑𝜃
𝑠𝑒𝑐2𝜃
1
2∫
𝑠𝑒𝑛2 𝜃
𝑐𝑜𝑠2 𝜃 1
𝑐𝑜𝑠2 𝜃
𝑑𝜃
1
2∫ 𝑠𝑒𝑛2𝜃 𝑑𝜃
1
2∫
1 − cos(2𝜃) 𝑑𝜃
2
1
4∫ 𝑑𝜃 −
1
4∫ cos(2𝜃) 𝑑𝜃
𝒖 = 2𝜃 ; 𝒅𝒖 = 2𝑑𝜃 ; 𝒅𝒖
𝟐= 𝑑𝜃
1
4. 𝜃 −
1
8∫ cos(𝑢). 𝑑𝑢
1
4. 𝜃 −
1
8 . sin(2𝜃) + 𝐶
1
4. 𝜃 −
1
4 . [sin(𝜃). cos( 𝜃) ] + 𝐶
𝒙 = 2. 𝑡𝑎𝑛 𝜃
𝑡𝑎𝑛−1 (𝑥
2) = 𝜽
1
4(tan−1 (
𝑥
2) − (
𝑥
√𝑥2 + 4 .
2
√𝑥2 + 4)) + 𝐶
𝟏
𝟒(𝒕𝒂𝒏−𝟏 (
𝒙
𝟐) − (
𝟐𝒙
𝒙𝟐 + 𝟒)) + 𝑪
2
𝑥
5)
∫𝑑𝑢
√𝑢2 + 𝑎2
Sol:
𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖ó𝑛: 𝑢 = 𝑎. tan 𝜃
𝑑𝑢 = 𝑎. 𝑠𝑒𝑐2𝜃 𝑑𝜃
∫𝑎. 𝑠𝑒𝑐2𝜃 𝑑𝜃
√(𝑎. 𝑡𝑎𝑛𝜃)2 + 𝑎2
𝑎 ∫𝑠𝑒𝑐2𝜃 𝑑𝜃
√𝑎2 (𝑡𝑎𝑛2𝜃 + 1)
𝑎
𝑎∫
𝑆𝑒𝑐2𝜃 𝑑𝜃
𝑠𝑒𝑐𝜃
∫ sec 𝜃 𝑑𝜃
ln|tan 𝜃 + sec 𝜃| + 𝐶
𝑢 = 𝑎. 𝑡𝑎𝑛𝜃
𝑢
𝑎 = tan 𝜃
ln |𝑢
𝑎 +
√𝑢2 + 𝑎2
𝑎| + 𝐶
ln |𝑎𝑢 +𝑎√𝑢2 +𝑎2
𝑎| + 𝐶
ln |𝑎(𝑢 + √𝑢2 + 𝑎2)
𝑎| + 𝐶
𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2| + 𝐶
6.1
6)
∫(4𝑥 − 2)
𝑥3 − 𝑥2 − 2𝑥𝑑𝑥
∫(4𝑥 − 2)
𝑥(𝑥2 − 𝑥 − 2)𝑑𝑥
∫(4𝑥 − 2)
𝑥(𝑥 − 2)(𝑥 + 1)𝑑𝑥
𝒂
𝒖
4𝑥 − 2 = (𝑥)(𝑥 − 2)(𝑥 + 1) [𝐴
𝑥+
𝐵
𝑥 − 2+
𝐶
𝑥 + 1]
4𝑥 − 2 = 𝐴(𝑥 − 2)(𝑥 + 1) + 𝐵(𝑥)(𝑥 + 1) + 𝐶(𝑥)(𝑥 − 2)
4𝑥 − 2 = 𝐴(𝑥2 + 𝑥 − 2𝑥 − 2) + 𝐵(𝑥2 + 𝑥) + 𝐶( 𝑥2 − 2𝑥)
4𝑥 − 2 = (𝐴𝑥2 − 𝐴𝑋 − 2𝐴) + (𝐵𝑥2 + 𝐵𝑋) + (𝐶𝑥2 − 2𝐶𝑋)
0 = 𝐴 + 𝐵 + 𝐶
4 = −𝐴 + 𝐵 − 2𝐶
2 = −2𝐴
−2 = −2𝐴
−2
−2= 𝐴 ; 1 = 𝐴
𝟎 = 𝑨 + 𝑩 + 𝑪
𝟒 = −𝑨 + 𝑩 − 𝟐𝑪
𝟒 = 𝟐𝑩 − 𝑪
𝑪 = 𝟐𝑩 − 𝟒
𝟒 = −𝟏 + 𝑩 − 𝟐(𝟐𝑩 − 𝟒)
𝟒 = −𝟏 + 𝑩 − 𝟒𝑩 + 𝟖
𝟒 = 𝟕 − 𝟑𝑩
−𝟑
−𝟑= 𝑩 ; 𝟏 = 𝑩
𝑯𝒂𝒍𝒍𝒂𝒏𝒅𝒐 𝑪:
𝑪 = 𝟐(𝟏) − 𝟒; 𝑪 = 𝟐 − 𝟒; 𝑪 = −𝟐
∫𝒅𝒙
𝒙+ ∫
𝒅𝒙
(𝒙 − 𝟐) − 𝟐 ∫
𝒅𝒙
(𝒙 − 𝟏)
ln[x] + ln[x − 2] − 2 ln[x + 1] + C
ln[(x). (x − 2)] − ln[(x + 1)2] + C
ln [𝑥(𝑥 − 2)
(𝑥 + 1)2] + 𝐶
𝟔. 𝟏
𝟏𝟖) ∫𝟒𝒙 + 𝟏
𝟐𝒙 + 𝟏𝒅𝒙
𝑢 = 2𝑥 𝑑𝑢 = 2𝑥 𝑙𝑜𝑔(2)𝑑𝑥
=1
𝑙𝑜𝑔 (2)∫
𝑢2 + 1
𝑢(𝑢 + 1)𝑑𝑢
=1
𝑙𝑜𝑔 (2)∫ (−
2
𝑢 + 1+
1
𝑢+ 1)𝑑𝑢
= −1
𝑙𝑜𝑔 (2)∫ 1𝑑𝑢 +
1
𝑙𝑜𝑔 (2)∫ −
1
𝑢𝑑𝑢 +
1
𝑙𝑜𝑔 (2)∫
1
𝑢 + 1𝑑𝑢
𝑠 = 𝑢 + 1 𝑑𝑠 = 𝑑𝑢
= −2 𝑙𝑜𝑔(𝑠)
𝑙𝑜𝑔 (2)+
1
𝑙𝑜𝑔 (2)∫ 1𝑑𝑢 +
1
𝑙𝑜𝑔 (2)∫
1
𝑢𝑑𝑢
= −2 𝑙𝑜𝑔 (𝑠)
𝑙𝑜𝑔 (2)+
𝑙𝑜𝑔 (𝑢)
𝑙𝑜𝑔 (2)+
1
𝑙𝑜𝑔 (2)∫ 1𝑑𝑢
= −2 𝑙𝑜𝑔 (𝑠)
𝑙𝑜𝑔 (2)+
𝑢
𝑙𝑜𝑔 (2)+
𝑙𝑜𝑔 (𝑢)
𝑙𝑜𝑔 (2)+ 𝐶
=𝑢 + 𝑙𝑜𝑔(2𝑥) − 2 𝑙𝑜𝑔 (2𝑥 + 1)
𝑙𝑜𝑔 (2)+ 𝐶
=𝟐𝒙 + 𝒙 𝒍𝒐𝒈(𝟐) − 𝟐 𝒍𝒐𝒈 (𝟐𝒙 + 𝟏)
𝒍𝒐𝒈 (𝟐)+ 𝑪
𝟔. 𝟐
𝟔) ∫ (𝒅𝒙
𝟗𝒙𝟒 + 𝒙𝟐)
= ∫ (1
𝑥2) − (
9
9𝑥4 + 1)
= ∫ (1
𝑥2) 𝑑𝑥 − ∫ (
1
9𝑥2 + 1)𝑑𝑥
𝑢 = 3𝑥 𝑑𝑢 = 3𝑑𝑥
= ∫1
𝑥2𝑑𝑥 − 3 ∫
1
𝑢2 + 1𝑑𝑢
= ∫1
𝑥2𝑑𝑥 − 3 tan−1 𝑢
= ∫1
𝑥2𝑑𝑥 − 3 tan−1 𝑢
−1
𝑥− 3 tan−1 𝑢 + 𝐶
−𝟏
𝒙− 𝟑 𝐭𝐚𝐧−𝟏(𝟑𝒙) + 𝑪
𝟏𝟖)
∫𝟐𝐱𝟐 + 𝟑𝐱 + 𝟒
𝐱𝟑 + 𝟒𝐱𝟐 + 𝟔𝐱 + 𝟒𝐝𝐱
= ∫ (2
x + 2−
1
x2 + 2x + 2 )
= 2 ∫1
x + 2dx − ∫
1
x2 + 2x + 2dx
= 2 ∫1
x + 2dx − ∫
1
(x + 1)2 + 1dx
u = x + 1 du = dx
= 2 ∫1
x + 2dx − ∫
1
u2 + 1du
=1
x + 2= s = x + 2 ds = dx
= 2 ∫1
sds − tan−1u
= 2 log(s) − tan−1(x + 1) + C
= 𝟐 𝒍𝒐𝒈 (𝒙 + 𝟐) − 𝒕𝒂𝒏−𝟏(𝒙 + 𝟏) + 𝑪