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Question Number 204264 by SANOGO last updated on 10/Feb/24

$$\underset{}{\int }ln(1-x^2)dx$$

Answered by witcher3 last updated on 10/Feb/24

$$\mathrm{by}\mathrm{part}\mathrm{u}=\ln (1-{\mathrm{x}}^2);{\mathrm{v}}^{\prime }\left(\mathrm{x}\right)=1$$$$=\mathrm{xln}(1-{\mathrm{x}}^2)-\int \frac{-2\mathrm{x}}{1-{\mathrm{x}}^2}.\mathrm{xdx}$$$$=\mathrm{xln}(1-{\mathrm{x}}^2)+2\int \frac{{\mathrm{x}}^2}{1-{\mathrm{x}}^2}\mathrm{dx}=\mathrm{xln}(1-{\mathrm{x}}^2)+2\int \frac{{\mathrm{x}}^2-1+\frac{1+\mathrm{x}}{2}+\frac{1-\mathrm{x}}{2}}{1-{\mathrm{x}}^2}\mathrm{dx}$$$$=\mathrm{xln}(1-{\mathrm{x}}^2)+2\int 1\mathrm{dx}+2\int \frac{1+\mathrm{x}+1-\mathrm{x}}{2(1-\mathrm{x})(1+\mathrm{x})}\mathrm{dx}$$$$=\mathrm{xln}(1-{\mathrm{x}}^2)+2\mathrm{x}+\int \frac{1}{(1-\mathrm{x})}+\frac{1}{1+\mathrm{x}}\mathrm{dx}$$$$=\mathrm{xln}(1-{\mathrm{x}}^2)+2\mathrm{x}+\ln \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)+\mathrm{c}$$

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