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. $none$

Answer 1

The correct option is B $4-\pi$

Put $e^x-1=t^2\Rightarrow e^{x}dx=2tdt;x=0\Rightarrow t=0,x=log5\Rightarrow t=2$
${\int }_0^{log5}\frac{e^x\sqrt{e^x-1}}{e^x-3}dx={\int }_0^2\frac{\sqrt{t^2}}{t^2+4}.2tdt=2{\int }_0^2\frac{t^2}{t^2+4}dt=2{\int }_0^2(1-\frac{4}{t^2+4})dt$
$=2{[t-\frac{4}{2}Ta{n}^{-1}\left(\frac{t}{2}\right)]}_0^2=2[2-2.\frac{\pi }{2}]=4-\pi$