Question

The value of the integral $\underset{0}{\overset{\ln 5}{\int }}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx$ is

• A. $3+2\pi$
• B. $4-\pi$
• C. $2+\pi$
• D. $\pi$

Answer 1

The correct option is B $4-\pi$

Putting $e^x-1=t^2,e^{x}dx=2tdt$ in the given integral, we have
$\underset{0}{\overset{\ln 5}{\int }}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx=2\underset{0}{\overset{2}{\int }}\frac{t^2}{t^2+4}dt=2\left(\underset{0}{\overset{2}{\int }}1dt-4\underset{0}{\overset{2}{\int }}\frac{dt}{t^2+4}\right)$
$=2\left[{(t-2{\tan }^{-1}\left(\frac{t}{2}\right))}_0^2\right]$
$=2\left[(2-2\times \frac{\pi }{4})\right]=4-\pi$