不定积分求解用第二类换元积分法求不定积分：∫2ex√（1-e2x）dx求不定积分：1、∫(x2+1)/［(x+1)2(x-1)］dx2、∫1/［(x2+1)(x2+x)］dx

Question

不定积分求解
用第二类换元积分法求不定积分：∫2ex√（1- e2x）dx求不定积分：1、∫(x2+1)/［(x+1)2(x-1)］dx2、∫1/［(x2+1)(x2+x)］dx 

Answer 1

Let y = e^x,dy = e^x dx
∫ 2e^x√[1 - e^(2x)] dx
= ∫ 2y√(1 - y²) * dy/y
= 2∫ √(1 - y²) dy,Let y = sinθ,dy = cosθ dθ
= 2∫ cos²θ dθ
= ∫ (1 + cos2θ) dθ
= θ + sinθcosθ + C
= arcsin(e^x) + √[1 - e^(2x)] e^x + C
------------------------------------------------------------------------------------
∫ (x² + 1)/[(x + 1)²(x - 1)] dx,设(x² + 1) = A/(x + 1)² + B/(x + 1) + C/(x - 1)
= ∫ {1/[2(x + 1)] - 1/(x + 1)² + 1/[2(x - 1)]} dx
= (1/2)ln|x + 1| + 1/(x + 1) + (1/2)ln|x - 1| + C
= 1/(x + 1) + (1/2)ln|(x + 1)/(x - 1)| + C
----------------------------------------------------------------------------
∫ dx/[(x² + 1)(x² + x)],设1 = (Ax + B)/(x² + 1) + C/(x + 1) + D/x
= ∫ {(-x - 1)/[2(x² + 1) - 1/[2(x + 1)] + 1/x} dx
= (1/2)∫ [-x/(x² + 1) - 1/(x² + 1)] dx - (1/2)ln|x + 1| + ln|x|
= (-1/4)ln(x² + 1) - (1/2) arctanx - (1/2)ln|x + 1| + ln|x| + C