Consider, I=∫√(e^x 1)dx e^x 1=t^2 e^xdx=2tdt (t^2+1)dx=2tdt dx=2tt^2+1 dt ∫2t^2t^2+1dt=∫2(t^2+1 1)t^2+1dt =2t 2tan^ 1(t^2+1)+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Global Maxima

Evaluate ∫√...

Question

Evaluate $\int \sqrt{e^{x} - 1} d x$

Open in App

Solution

consider, $I = \int \sqrt{e^{x} - 1} d x$

$e^{x} - 1 = t^{2}$
$e^{x} d x = 2 t d t$

$(t^{2} + 1) d x = 2 t d t$

$d x = \frac{2 t}{t^{2} + 1} d t$

$\int \frac{2 t^{2}}{t^{2} + 1} d t = \int \frac{2 (t^{2} + 1 - 1)}{t^{2} + 1} d t$
$= 2 t - 2 {tan}^{- 1} (t^{2} + 1) + C$

$= 2 \sqrt{e^{x} - 1} - 2 {tan}^{- 1} (e^{x}) + c$

Suggest Corrections