Question

求不定积分，谁会做啊，求过程谢谢！

Answer 1

追答

Take the integral:
integral 1/((x+1) sqrt(x^2+1)) dx
For the integrand 1/((x+1) sqrt(x^2+1)), substitute x = tan(u) and dx = sec^2(u) du. Then sqrt(x^2+1) = sqrt(tan^2(u)+1) = sec(u) and u = tan^(-1)(x):
= integral (sec(u))/(tan(u)+1) du
For the integrand (sec(u))/(tan(u)+1), substitute s = tan(u/2) and ds = 1/2 du sec^2(u/2). Then transform the integrand using the substitutions sin(u) = (2 s)/(s^2+1), cos(u) = (1-s^2)/(s^2+1) and du = (2 ds)/(s^2+1):
= integral 2/((1-s^2) (1-(2 s)/(s^2-1))) ds
Simplify the integrand 2/((1-s^2) (1-(2 s)/(s^2-1))) to get 2/(-s^2+2 s+1):
= integral 2/(-s^2+2 s+1) ds
Factor out constants:
= 2 integral 1/(-s^2+2 s+1) ds
For the integrand 1/(-s^2+2 s+1), complete the square:
= 2 integral 1/(2-(s-1)^2) ds
For the integrand 1/(2-(s-1)^2), substitute p = s-1 and dp = ds:
= 2 integral 1/(2-p^2) dp
Factor 2 from the denominator:
= 2 integral 1/(2 (1-p^2/2)) dp
Factor out constants:
= integral 1/(1-p^2/2) dp
For the integrand 1/(1-p^2/2), substitute w = p/sqrt(2) and dw = 1/sqrt(2) dp:
= sqrt(2) integral 1/(1-w^2) dw
The integral of 1/(1-w^2) is tanh^(-1)(w):
= sqrt(2) tanh^(-1)(w)+constant
Substitute back for w = p/sqrt(2):
= sqrt(2) tanh^(-1)(p/sqrt(2))+constant
Substitute back for p = s-1:
= sqrt(2) tanh^(-1)((s-1)/sqrt(2))+constant
Substitute back for s = tan(u/2):
= sqrt(2) tanh^(-1)((tan(u/2)-1)/sqrt(2))+constant
Substitute back for u = tan^(-1)(x):
= sqrt(2) tanh^(-1)((-sqrt(x^2+1)+x-1)/(sqrt(2) (sqrt(x^2+1)+1)))+constant
Which is equivalent for restricted x values to:
Answer: |
| = (log(x+1)-log(sqrt(2) sqrt(x^2+1)-x+1))/sqrt(2)+constant

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看不懂

追答

图片？

这些英文猜猜意思就行了

不用过多了解

追问

图片发的什么，我手机上看不清。就说这题怎么做就行了，不用那么复杂

追答

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谢谢了

Answer 2

令x=tant

追问

过程。。

Answer 3

追答

括号里应该是加上派/4

追问

消去根号应该加上绝对值吧

追答

额