Question

求一个不定积分∫2x/ [(x+1)(x²+1)²]dx

求∫2x/ [(x+1)(x²+1)²]dx 给出详细算法

Answer 1

原式=∫[（2x-2+3）/(x^2-2x+2)]dx
=∫[（2x-2）/(x^2-2x+2)]dx+∫[3/(x^2-2x+2)]dx
=∫[1/(x^2-2x+2)]d(x^2-2x+2)+3∫{1/[(x-1)^2+1]}d(x-1)
=ln(x^2-2x+2)+3arctan(x-1)+C

追问

分子是2x 分母是x+1和（x²+1)²的乘积