xln[(1+x)/(1-x)]不定积分

Question

xln[(1+x)/(1-x)]不定积分

Answer 1

∫ xln[(1 + x)/(1 - x)] dx
= ∫ 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