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Question Number 141560 by mnjuly1970 last updated on 20/May/21

$$$$$$........Nice....Calculus\left(I\right)......$$$$Evaluate::$$$$\mathrm{I}:={\int }_0^{ln\left(2\right)}\frac{x}{e^x+2e^{-x}-2}dx=?$$$$......$$

Answered by mindispower last updated on 20/May/21

$$={\int }_0^{ln\left(2\right)}\frac{{xe}^x}{{(e^x-1)}^2+1}dx$$$$u=e^x-1$$$$={\int }_0^1\frac{ln(1+u)}{u^2+1}du,u=tg\left(t\right)$$$$={\int }_0^{\frac{\pi }{4}}ln(1+tg\left(t\right))$$$$={\int }_0^{\frac{\pi }{4}}ln\left(\frac{\sqrt{2}sin(\frac{\pi }{4}+u)}{cos\left(u\right)}\right)du$$$$=\frac{\pi }{4}ln\left(\sqrt{2}\right)+{\int }_0^{\frac{\pi }{4}}ln\left(sin(\frac{\pi }{4}+u)\right)du-{\int }_0^{\frac{\pi }{4}}ln\left(cos\left(u\right)\right)du$$$${\int }_0^{\frac{\pi }{4}}ln\left(sin(u+\frac{\pi }{4})\right)du={\int }_0^{\frac{\pi }{4}}ln\left(cos\left(u\right)\right)du\mid u\to \frac{\pi }{4}-u$$$$\iff {\int }_0^{ln\left(2\right)}\frac{x}{e^x+2e^{-x}-2}dx=\frac{\pi }{4}ln\left(\sqrt{2}\right)=\frac{\pi ln\left(2\right)}{8}$$$$$$

Commented by mnjuly1970 last updated on 20/May/21

$$gratefulsirpower...$$

Commented by mindispower last updated on 20/May/21

$$pleasursir$$

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