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Question Number 144716 by mathdanisur last updated on 28/Jun/21

$$\int \left(\frac{dx}{e^{2x}+1}\right)=?$$

Answered by imjagoll last updated on 28/Jun/21

$$\mathrm{let}{\mathrm{e}}^{\mathrm{x}}=\mu \Rightarrow \mathrm{dx}=\frac{\mathrm{d}\mu }{{\mathrm{e}}^{\mathrm{x}}}=\frac{\mathrm{d}\mu }{\mu }$$$$\mathrm{I}=\int \left(\frac{1}{{\mu }^2+1}\right)\left(\frac{\mathrm{d}\mu }{\mu }\right)$$$$\frac{1}{\mu ({\mu }^2+1)}=\frac{\mathrm{a}}{\mu }+\frac{\mathrm{b}\mu +\mathrm{c}}{{\mu }^2+1}$$$$1=(\mathrm{a}+\mathrm{b}){\mu }^2+\mathrm{c}\mu +\mathrm{a}$$$$\to \mu =0\to \mathrm{a}=1$$$$\mu =1\to 1=1+\mathrm{b}+\mathrm{c}+1$$$$\to \mathrm{b}+\mathrm{c}=-1$$$$\mu =-1\to 1=1+\mathrm{b}-\mathrm{c}+1$$$$\to \mathrm{b}-\mathrm{c}=-1\{\begin{array}{l}\mathrm{b}=-1\\ \mathrm{c}=0\end{array}$$$$\mathrm{I}=\int (\frac{1}{\mu }-\frac{\mu }{{\mu }^2+1})\mathrm{d}\mu$$$$\mathrm{I}=\ln \mu -\frac{1}{2}\int \frac{\mathrm{d}({\mu }^2+1)}{{\mu }^2+1}$$$$\mathrm{I}=\ln \mu -\frac{1}{2}\ln ({\mu }^2+1)+\mathrm{c}$$$$\mathrm{I}=\ln {\mathrm{e}}^{\mathrm{x}}-\frac{1}{2}\ln ({\mathrm{e}}^{2\mathrm{x}}+1)+\mathrm{c}$$$$\mathrm{I}=\mathrm{x}-\frac{1}{2}\ln ({\mathrm{e}}^{2\mathrm{x}}+1)+\mathrm{c}$$

Commented by mathdanisur last updated on 28/Jun/21

$$thanksSircool$$

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