Question

$\underset{0}{\overset{\log 5}{\int }}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx=$

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

Answer 1

Answer: (B)

Put $e^x-1=r^2$
$\therefore e^{x}dx=2tdt$
when $x=0,t=0$
when $x=log5,r^2=e^{log5}-1=5-1=4$
$\therefore t=2$
$\therefore$ given $integral=\int limit{s}_0^2\frac{t\cdot 2tdt}{t^2+1+3}$
$=\underset{0}{\overset{2}{\int }}\frac{2t^2}{t^2+4}dt$
$=2\underset{0}{\overset{2}{\int }}\frac{t^2+4-4}{t^2+4}dt$
$=2\underset{0}{\overset{2}{\int }}dt-8\underset{0}{\overset{2}{\int }}\frac{dt}{t^2+4}$
$=2{\left|t\right|}_0^2-\frac{8}{2}{\left|ta{n}^{-1}\frac{t}{2}\right|}_0^2$
$=4-4\left[ta{n}^{-1}1-ta{n}^{-1}0\right]$
$=4-4\frac{\pi }{4}=4-\pi$