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Question Number 148564 by mathmax by abdo last updated on 29/Jul/21

$$\mathrm{calculate}{\int }_1^2\frac{\mathrm{logx}}{1+\mathrm{x}}\mathrm{dx}$$

Answered by Kamel last updated on 29/Jul/21

$$$$$$\mathrm{\Omega }={\int }_1^2\frac{Ln\left(x\right)}{1+x}dx=-{\int }_{\frac{1}{2}}^1\frac{Ln\left(x\right)}{x(1+x)}dx$$$$=\frac{1}{2}{Ln}^2\left(2\right)+Ln\left(2\right)Ln\left(\frac{3}{2}\right)-{\int }_{\frac{1}{2}}^1\frac{Ln(1+x)}{x}dx$$$$=Ln\left(2\right)Ln\left(3\right)-\frac{1}{2}{Ln}^2\left(2\right)+{Li}_2(-1)-{Li}_2(-\frac{1}{2})$$$$\therefore {\int }_1^2\frac{Ln\left(x\right)}{1+x}dx=Ln\left(2\right)Ln\left(3\right)-\frac{1}{2}{Ln}^2\left(2\right)-\frac{{\pi }^2}{12}-{Li}_2(-\frac{1}{2})$$$$$$

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