ln(1+x)\(1+x^2)的不定积分

网友回答

∫ ln[(1+x)/(1+x²)] dx
= xln[(1+x)/(1+x²)] - ∫ x dln[(1+x)/(1+x²)]
= xln[(1+x)/(1+x²)] - ∫ x(1+x²)/(1+x) * [(1+x²)-(1+x)(2x)]/(1+x²)² dx
= xln[(1+x)/(1+x²)] + ∫ x(x²+2x-1)/[(1+x)(1+x²)] dx
= xln[(1+x)/(1+x²)] + ∫ [-2/(1+x²)+1/(1+x)+1] dx
= xln[(1+x)/(1+x²)] - 2∫ dx/(1+x²) + ∫ dx/(1+x) + ∫ dx
= xln[(1+x)/(1+x²)] - 2arctanx + ln|x|
Note：x(x²+2x-1)/[(1+x)(1+x²)] = A/(1+x²) + B/(x+1) + C
x³+2x²-x = A(x+1) + B(1+x²) + C(1+x²)(x+1)
Put x = -1,2 = B(2) => B = 1Put x = 0,0 = A + 1 + C => C = -A-1
Put x = 1,2 = 2A + 2 + 4C => C = -A/2
-A-1 = -A/2
2A+2 = A
A = -2C = -(-2)/2 = 1
∴x(x²+2x-1)/[(1+x)(1+x²)] = -2/(1+x²) + 1/(x+1) + 1
======以下答案可供参考======
供参考答案1:
Sln(1+x^2)/(1+x)dx
=xln(1+x^2)/(1+x)-Sxdln(1+x^2)/(1+x)
=xln(1+x^2)/(1+x)-Sx*(2x(1+x)-(1+x^2))/(1+x)^2)dx
=xln(1+x^2)/(1+x)-S((x^3+2x^2+x-2(x+1)+2)/(1+x)^2)dx
=xln(1+x^2)/(1+x)-S(x-2/(x+1)+2/(x+1)^2)dx
=xln(1+x^2)/(1+x)-1/2*x^2+2ln(x+1)+2/(x+1)+c
对数的真数是（x+1)分子（x^2+1)吧，(x^2+1)是分子，(x+1)是分母