Question

Integrate the following with respect to $x$.
$\sqrt{e^x+1}$

Answer 1

Consider the given integral.

$I=\int \sqrt{e^x+1}dx$

Let $t^2=e^x+1$

On differentiating both sides w.r.t $x$, we get

$2tdt=e^{x}dx$

Therefore,

$I=\int \sqrt{t^2}\frac{\left(2tdt\right)}{(t^2-1)}$

$I=2\int \frac{t^2}{(t^2-1)}dt$

$I=2\int \frac{t^2-1+1}{(t^2-1)}dt$

$I=2\int 1dt+2\int \frac{1}{(t^2-1)}dt$

$I=2t+2\left(\frac{1}{2}\log \left(\frac{t-1}{t+1}\right)\right)+C$

On putting the value of $t$, we have,

$I=2\sqrt{e^x+1}+\log \left(\frac{\sqrt{e^x+1}-1}{\sqrt{e^x+1}+1}\right)+C$

Hence, this is the answer.