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Question Number 137206 by JulioCesar last updated on 31/Mar/21
Answered by bemath last updated on 31/Mar/21
byparts{u=ln(x−1x+1)du=2(x−1)(x+1)dxv=xI=xln(x−1x+1)−∫2x(x−1)(x+1)dxI=xln(x−1x+1)−[∫1x−1dx+∫1x+1dx]I=xln(x−1x+1)−ln(x2−1)+CI=(x−1)ln(x−1)+(1−x)ln(x+1)+C
Answered by Ñï= last updated on 31/Mar/21
∫lnx−1x+1dx=∫ln(x−1)dx−∫ln(x+1)dx=(x−1)ln(x−1)−(x−1)−(x+1)ln(x+1)+(x+1)+C=(x−1)ln(x−1)−(x+1)ln(x+1)+C
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