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ln-1-x-2-dx-




Question Number 204264 by SANOGO last updated on 10/Feb/24
∫_ ln(1−x^2 )dx
$$\underset{} {\int}{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$
Answered by witcher3 last updated on 10/Feb/24
by part u=ln(1−x^2 );v′(x)=1  =xln(1−x^2 )−∫((−2x)/(1−x^2 )).xdx  =xln(1−x^2 )+2∫(x^2 /(1−x^2 ))dx=xln(1−x^2 )+2∫((x^2 −1+((1+x)/2)+((1−x)/2))/(1−x^2 ))dx  =xln(1−x^2 )+2∫1dx+2∫((1+x+1−x)/(2(1−x)(1+x)))dx  =xln(1−x^2 )+2x+∫(1/((1−x)))+(1/(1+x))dx  =xln(1−x^2 )+2x+ln(((1+x)/(1−x)))+c
$$\mathrm{by}\:\mathrm{part}\:\mathrm{u}=\mathrm{ln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right);\mathrm{v}'\left(\mathrm{x}\right)=\mathrm{1} \\ $$$$=\mathrm{xln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)−\int\frac{−\mathrm{2x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }.\mathrm{xdx} \\ $$$$=\mathrm{xln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)+\mathrm{2}\int\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\mathrm{xln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)+\mathrm{2}\int\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{1}+\frac{\mathrm{1}+\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{1}−\mathrm{x}}{\mathrm{2}}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\mathrm{xln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)+\mathrm{2}\int\mathrm{1dx}+\mathrm{2}\int\frac{\mathrm{1}+\mathrm{x}+\mathrm{1}−\mathrm{x}}{\mathrm{2}\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}\right)}\mathrm{dx} \\ $$$$=\mathrm{xln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)+\mathrm{2x}+\int\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{xln}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)+\mathrm{2x}+\mathrm{ln}\left(\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}\right)+\mathrm{c} \\ $$

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