∫xln(1-x)/(1+x)dx 的不定积分?
问题描述：
∫xln(1-x)/(1+x)dx 的不定积分?
答
= ∫ ln[(1 + x)/(1 - x)] d(x²/2)
= (1/2)x²ln[(1 + x)/(1 - x)] - (1/2)∫ x² d[ln(1 + x)/(1 - x)]
= (1/2)x²ln[(1 + x)/(1 - x)] - (1/2)∫ x² * 2/(1 - x²) dx
= (1/2)x²ln[(1 + x)/(1 - x)] + ∫ [(1 - x²) - 1]/(1 - x²) dx
= (1/2)x²ln[(1 + x)/(1 - x)] + ∫ dx - ∫ dx/(1 - x²)
= (1/2)x²ln[(1 + x)/(1 - x)] + x + (1/2)ln|(1 - x)/(1 + x)| + C解决了吗？解决了麻烦采纳一下，谢谢了
∫ln[(1+x)/(1-x)] dx=x*ln[(1+x)/(1-x)]-∫x d{ln[(1+x)/(1-x)]},分部积分法=xln[(1+x)/(1-x)]-∫x*-2/(x^2-1) dx,对ln[(1+x)/(1-x)]求微分=xln[(1+x)/(1-x)]+2∫x/(x^2-1) dx,令t=x^2-1,dt=2x dx→dx=dt/(2x)=xln[...ln(1+x)/(1-x)前面还有个x呀
= ∫ ln[(1 + x)/(1 - x)] d(x²/2)
= (1/2)x²ln[(1 + x)/(1 - x)] - (1/2)∫ x² d[ln(1 + x)/(1 - x)]
= (1/2)x²ln[(1 + x)/(1 - x)] - (1/2)∫ x² * 2/(1 - x²) dx
= (1/2)x²ln[(1 + x)/(1 - x)] + ∫ [(1 - x²) - 1]/(1 - x²) dx
= (1/2)x²ln[(1 + x)/(1 - x)] + ∫ dx - ∫ dx/(1 - x²)
= (1/2)x²ln[(1 + x)/(1 - x)] + x + (1/2)ln|(1 - x)/(1 + x)| + C解决了吗？解决了麻烦采纳一下，谢谢了