Question

${\int }_0^{log5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx=$

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

Answer 1

Answer: (B)

$4-\pi$

Let $I=\int \frac{e^x\sqrt{e^x-1}}{e^x+3}$
Substitute $t=e^x\Rightarrow dt=e^{x}dx$, we get
$I=\int \frac{\sqrt{t-1}}{t+3}dt$
Substitute $u=\sqrt{t-1}\Rightarrow du=\frac{1}{2\sqrt{t-1}}dt$, we get
$I=2\int \frac{u^2}{u^2+4}du=2\int (1-\frac{4}{u^2+4})du=2\int du-2\int \frac{4}{u^2+4}du=2u-4{tan}^{-1}\left(\frac{u}{2}\right)=2\sqrt{t-1}-4{tan}^{-1}\left(\frac{\sqrt{t-1}}{2}\right)=2\sqrt{e^x-1}-4{tan}^{-1}\left(\frac{\sqrt{e^x-1}}{2}\right)$
Therefore,
${\int }_0^{\log 5}Idx={[2\sqrt{e^x-1}-4{tan}^{-1}\left(\frac{\sqrt{e^x-1}}{2}\right)]}_0^{\log 5}=4-\pi$