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Question Number 126316 by benjo_mathlover last updated on 19/Dec/20

$$\underset{\sqrt{2}}{\stackrel{2e^{\sqrt{2}}}{\int }}\ln \left(\frac{e^x+e^{-x}}{9\sqrt{x}}\right)dx?$$

Commented by liberty last updated on 19/Dec/20

$$\int (\ln \left(e^x(1+e^{-2x})\right)-\ln 9-\frac{1}{2}\ln x)dx=$$$$\int x+\ln (1+e^{-2x})-\ln 9-\frac{1}{2}\ln xdx=$$$$\frac{1}{2}{x}^2-x\ln 9-\frac{1}{2}(x\ln x-x)+\int \ln (1+e^{-2x})dx=$$$$\frac{1}{2}{x}^2+\frac{1}{2}x-x\ln 9\sqrt{x}+\int \ln (1+e^{-2x})dx=$$$$$$

Commented by Olaf last updated on 19/Dec/20

$$\mathrm{Let}u=e^{-x}$$$$\int \ln (1+e^{-2x})dx=-\int \frac{\ln (1+u^2)}{u}du$$$$\frac{1}{1+u}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{u}^n$$$$\frac{1}{1+u^2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{u}^{2n}$$$$\frac{2u}{1+u^2}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{u}^{2n+1}$$$$\ln (1+u^2)=2\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{u^{2n+2}}{2n+2}$$$$-\frac{\ln (1+u^2)}{u}=-2\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{u^{2n+1}}{2n+2}$$$$-\int \frac{\ln (1+u^2)}{u}du=-2\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{u^{2n+2}}{{(2n+2)}^2}$$$$-\int \frac{\ln (1+u^2)}{u}du=\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{u^{2n}}{n^2}$$$$\int \ln (1+e^{-2x})dx=\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{e^{-2nx}}{n^2}$$

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