Question

Evaluate : ${\int }_0^{{\log }_{e}5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx$

Answer 1

$\to$ Given to solve ${\int }_0^{ln5}\frac{e^x\sqrt{e^x-1}}{e^x+5}dx$

let $e^x-1=t^2$

$\therefore$ The new limits of integration are

$t_1=e^0-1=0,t_2=e^{ln5}-1=4$

and $e^{x}dx=2tdt$

$\Rightarrow {\int }_0^4\frac{2t^{2}dt}{t^2+4}\Rightarrow 2{\int }_0^4\frac{t^{2}dt}{t^2+4}$

$\Rightarrow 2\left[{\int }_0^4\frac{(t^2+4)-4}{t^2+4}dt\right]$

$\Rightarrow 2\left[{\int }_0^{4}ldt-{\int }_0^4\frac{4}{t^2+4}dt\right]$

$\Rightarrow 2[4-\frac{4}{4}{\tan }^{-1}{\left(\frac{t}{2}\right)}^4]$

$\Rightarrow 2(4-{\tan }^{-1}\left(2\right))$

$=8-2{\tan }^{-1}2$

$\therefore {\int }_0^{ln5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx=8-2{\tan }^{-1}2$