Question

i={ 1 /( e^x + e-^x dx
integer​

Answer 1

$\sf\large\red{\underline{\underline{Appropriate\; Question}}}:$

• $\bf\displaystyle\int\dfrac{1}{e^x + e^{-x}}dx$

$\sf\large\red{\underline{\underline{Solution}}}:$

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

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

 $\boxed{\red\bigstar e^{-x} = \dfrac{1}{e^x}}$

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

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

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

Now put $e^{2x}$ = t

differentiating both sides w.r.t x,

$2e^{2x}\;dx = dt$

$e^{2x}\;dx = \dfrac{dt}{2}$

so,

$= \displaystyle\int\dfrac{dt}{t^2+1}$

We know that, $\red\bigstar\displaystyle\int\dfrac{1}{x^2+1} dx = \tan^{-1}\frac{x}{a}+ c$

$= \sf\tan^{-1}\frac{t}{1} + c$

Putting t = e^2x

$= \displaystyle\tan^{-1} e^{2x} + c$