Question

intergrate
dx divided by
$e {}^{x \: } \: + e {}^{ - x} \:$
​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

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$\boxed{ \red{ \bf \: \tt \: \int \dfrac{dx}{ {x}^{2} + 1 } = {tan}^{ - 1} x + c}}$

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Concept used :-

• Method of Substitution

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$\large\underline\purple{\bold{Solution :- }}$

$\tt \:Let \: I \: =  \int \dfrac{dx}{ {e}^{x} + {e}^{ - x} }$

$= \tt \: \int \dfrac{dx}{{e}^{x} + \dfrac{1}{{e}^{x}} }$

$= \tt \: \int \dfrac{{e}^{x}}{{e}^{2x} + 1}$

$\tt \:  Put \: {e}^{x} \: = \: t \: \: \: \: \: \: \: \: \: \: \\ \tt\implies \:\dfrac{d}{dx} {e}^{x} = \dfrac{dt}{dx} \\ \tt\implies \:{e}^{x} = \dfrac{dt}{dx} \: \: \: \: \: \\ \bf\implies \:{e}^{x}dx = dt$

$\tt \: So, \: I \: = \tt \: \int \dfrac{dt}{ {t}^{2} + 1}$

$\bf\implies \:I = {tan}^{ - 1} t \: + \: c$

$\bf\implies \:I = {tan}^{ - 1} ({e}^{x}) \: + \: c$

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Answer 2

Given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{e^x+e^{-x}}$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{e^x+\dfrac{1}{e^x}}$

$\displaystyle\longrightarrow I=\int\dfrac{dx}{\left(\dfrac{e^{2x}+1}{e^x}\right)}$

$\displaystyle\longrightarrow I=\int\dfrac{e^x}{e^{2x}+1}\ dx$

$\displaystyle\longrightarrow I=\int\dfrac{e^x\ dx}{1+\left(e^x\right)^2}\quad\quad\dots(1)$

Put,

$\longrightarrow u=e^x$

$\longrightarrow du=e^x\ dx$

Then (1) becomes,

$\displaystyle\longrightarrow I=\int\dfrac{du}{1+u^2}$

$\displaystyle\longrightarrow I=\tan^{-1}u+C$

$\displaystyle\longrightarrow\underline{\underline{I=\tan^{-1}\left(e^x\right)+C}}$