求积分∫（1+x^2）⼀(x+1)*(1-x)^2 dx 多谢

设1 + x² = A/(x + 1) + B/(1 - x)² + C/(1 - x)
1 + x² = A(1 - x)² + B(x + 1) + C(1 - x)(x + 1)
1 + x² = A(1 - 2x + x²) + B(x + 1) + C(1 - x²)
1 + x² = (A - C)x² + (- 2A + B)x + (A + B + C)
则
A - C = 1 ==> C = A - 1
- 2A + B = 0 ==> B = 2A
A + B + C = 1 ==> A + (2A) + (A - 1) = 1
4A = 2 ==> A = 1/2，B = 1，C = - 1/2
∴
∫ (1 + x²)/[(x + 1)(1 - x)²] dx
= (1/2)∫ dx/(x + 1) + ∫ dx/(1 - x)² - (1/2)∫ dx/(1 - x)
= (1/2)∫ d(x + 1)/(x + 1) - ∫ d(1 - x)/(1 - x)² + (1/2)∫ d(1 - x)/(1 - x)
= (1/2)ln|1 + x| + 1/(1 - x) + (1/2)ln|1 - x| + C
= 1/(1 - x) + (1/2)ln|1 - x²| + C