Question

◆ Integration ◆

Integrate ( w.r.t. x ) -
$\sqrt{ {e}^{x} - 1} \: .dx$

Answer 1

Hey Starry :p

$\int\limits \sqrt{e^x - 1} dx$ <--- Question √√

$Substitute \ e^x = u \ \ -\ \textgreater \ du = e^xdx$

Now,
 $\int\limits { \frac{\sqrt{u-1}}{u} } du$

Again substitute s = √(u-1) ;
-> [tex]2 \int\limits { \frac{s^2}{s^2+1}} ds \\ \\ = 2\int\limits {(1 - \frac{1}{s^2 + 1}) } dx \\ \\ = 2s - 2tan^{-1}s + c[/tex]

=> The required integral is :->
 --> $2 \sqrt{e^x - 1} - 2tan^{-1}(\sqrt{e^x - 1}) + c$

Answer 2

this is Ur required result in attachment