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Question Number 90802 by john santu last updated on 26/Apr/20

$$1)\underset{\ln 2}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{{\mathrm{e}}^x-1}}dx$$$$2)\underset{\ln 2}{\stackrel{\mathrm{\infty }}{\int }}\frac{1}{e^x-1}dx$$

Commented by john santu last updated on 26/Apr/20

$$1)u=\sqrt{e^x-1}\Rightarrow dx=\frac{2udu}{e^x}$$$$\int \underset{1}{\stackrel{\mathrm{\infty }}{}}\frac{2u}{u}\frac{1}{e^x}du=\underset{1}{\stackrel{\mathrm{\infty }}{\int }}2\frac{du}{u^2+1}$$$${={\tan }^{-1}\left(u\right)]}_1^{\mathrm{\infty }}=2(\frac{\pi }{2}-\frac{\pi }{4})=\frac{\pi }{2}$$$$2)\underset{\ln 2}{\stackrel{\mathrm{\infty }}{\int }}\frac{e^{-x}}{(e^x-1)e^{-x}}dx=$$$$\text{Missing left or extra right}$$$$=0-\ln \left(\frac{1}{2}\right)$$$$=\ln 2$$

Commented by mathmax by abdo last updated on 26/Apr/20

$$2){\int }_{ln2}^{\mathrm{\infty }}\frac{dx}{e^x-1}{=}_{e^x=t}{\int }_2^{\mathrm{\infty }}\frac{dt}{t(t-1)}$$$$={\int }_2^{\mathrm{\infty }}(\frac{1}{t-1}-\frac{1}{t})dt={[ln\mid \frac{t-1}{t}\mid ]}_2^{+\mathrm{\infty }}=-ln\left(\frac{1}{2}\right)=ln\left(2\right)$$

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