Question

∫ln(1-x^2)dx 怎么求？

Answer 1

1-x^2>0
-1<x<1
∫ln(1-x^2)dx
=xln(1-x^2)- ∫x d[ln(1-x^2)]
=xln(1-x^2)- ∫x[-2x/(1-x^2)] dx
=xln(1-x^2)+ 2∫x^2/(1-x^2) dx
=xln(1-x^2)+ 2∫[-1+1/(1-x^2)]dx
=xln(1-x^2)+ 2[-x+1/2∫[1/(1-x)+1/(1+x)] dx ]
=xln(1-x^2)-2x-ln(1-x)+ln(1+x)+c
希望帮助到你，望采纳，谢谢~

Answer 2

∫ln(1-x^2)dx
=xln(1-x²)-∫xdln(1-x^2)
=xln(1-x²)-∫x*1/(1-x²)*(-2x)dx
=xln(1-x²)-2∫x²/(x²-1)dx
=xln(1-x²)-2∫(x²-1+1)/(x²-1)dx
=xln(1-x²)-2∫[1+1/(x+1)(x-1)]dx
=xln(1-x²)-∫[2+1/(x-1)-1/(x+1)]dx
=xln(1-x²)-2x-ln|x-1|+ln|x+1|+C

Answer 3

分部积分