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Question Number 137206 by JulioCesar last updated on 31/Mar/21

Answered by bemath last updated on 31/Mar/21

$$\mathrm{by}\mathrm{parts}\{\begin{array}{l}\mathrm{u}=\ln \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)\mathrm{du}=\frac{2}{(\mathrm{x}-1)(\mathrm{x}+1)}\mathrm{dx}\\ \mathrm{v}=\mathrm{x}\end{array}$$$$\mathrm{I}=\mathrm{x}\ln \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)-\int \frac{2\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}+1)}\mathrm{dx}$$$$\mathrm{I}=\mathrm{x}\ln \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)-[\int \frac{1}{\mathrm{x}-1}\mathrm{dx}+\int \frac{1}{\mathrm{x}+1}\mathrm{dx}]$$$$\mathrm{I}=\mathrm{x}\ln \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)-\ln ({\mathrm{x}}^2-1)+\mathrm{C}$$$$\mathrm{I}=(\mathrm{x}-1)\ln (\mathrm{x}-1)+(1-\mathrm{x})\ln (\mathrm{x}+1)+\mathrm{C}$$

Answered by Ñï= last updated on 31/Mar/21

$$\int ln\frac{x-1}{x+1}dx=\int ln(x-1)dx-\int ln(x+1)dx$$$$=(x-1)ln(x-1)-(x-1)-(x+1)ln(x+1)+(x+1)+C$$$$=(x-1)ln(x-1)-(x+1)ln(x+1)+C$$

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