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Question Number 140866 by liberty last updated on 13/May/21

$$\underset{0}{\stackrel{1}{\int }}\frac{\ln 2-\ln (1+{\mathrm{x}}^2)}{1-\mathrm{x}}\mathrm{dx}=?$$

Answered by mindispower last updated on 15/May/21

$$=\underset{x\to 1}{\lim }{\int }_0^x\frac{ln\left(2\right)-ln(1+t^2)}{1-t}dt$$$$=\underset{x\to 1}{\lim }-ln\left(2\right)ln(1-x)+ln(1-x)ln(1+x^2)-{\int }_0^x\frac{ln(1-t)2t}{1+t^2}dt$$$$=\underset{x\to 1}{\lim }ln(1-x).ln\left(\frac{1+x^2}{2}\right)=\underset{x\to 1}{\lim }ln(1-\frac{1-x^2}{2})ln(1-x)=0$$$$=-{\int }_0^1\frac{ln(1-x).2xdx}{1+x^2}=-A$$$$letB={\int }_0^1\frac{ln(1+x).2xdx}{1+x^2}$$$$A+B={\int }_0^1\frac{ln(1-x^2)}{1+x^2}.d\left(x^2\right)$$$$={\int }_0^1\frac{ln(1-t)}{1+t}dt={\int }_0^1\frac{ln\left(t\right)}{2-t}dt$$$$={\int }_0^{\frac{1}{2}}\frac{ln\left(2w\right)}{1-w}.dw={ln}^2\left(2\right)+{\int }_0^{\frac{1}{2}}\frac{ln\left(w\right)}{1-w}dw$$$$={ln}^2\left(2\right)+{\int }_1^{\frac{1}{2}}-\frac{ln(1-t)}{t}={ln}^2\left(2\right)+{li}_2\left(\frac{1}{2}\right)-{li}_2\left(1\right)$$$$A-B={\int }_0^1\frac{ln\left(\frac{1-t}{1+t}\right).2tdt}{1+t^2},t=\frac{1-x}{1+x}$$$$dt=\frac{-2}{{(1+x)}^2}dx$$$$=4{\int }_0^1\frac{ln\left(x\right)}{2+2x^2}.\frac{1-x}{1+x}dx$$$${\int }_0^1\frac{(1-x)ln\left(x\right)}{(1+x)(1+x^2)}dx$$$$={\int }_0^1\frac{ln\left(x\right)}{1+x}-\frac{x}{1+x^2}ln\left(x\right)dx$$$$=\frac{3}{4}{\int }_0^1\frac{ln\left(x\right)}{1+x}dx=\frac{3}{4}.-{\int }_0^1\frac{ln(1-(-x))}{-x}.d(-x)$$$$=-{\int }_0^{-1}\frac{ln(1-t)}{t}dt=\frac{-3}{4}{li}_2(-1)$$

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