Question

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

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

Answer 1

Answer: (B)

Put $e^{x}- 1 = r^{2} $
$\therefore e^{x} dx=2t dt$
when $ x = 0, t = 0$
when $x = log 5, r^{2}=e^{log 5}-1=5-1=4$
$\therefore t=2$
$\therefore $ given $integral=\int limits_{0}^{2} \frac{t\cdot2t dt}{t^{2}+1+3}$
$=\int\limits_{0}^{2} \frac{2t^{2}}{t^{2}+4}dt$
$=2\int\limits_{0}^{2} \frac{t^{2}+4-4}{t^{2}+4} dt $
$=2 \int\limits_{0}^{2} dt -8 \int\limits_{0}^{2}\frac{dt}{t^{2}+4}$
$=2\left|t\right|_{0}^{2}-\frac{8}{2}\left|tan^{-1}\frac{t}{2}\right|_{0}^{2}$
$=4-4\left[tan^{-1} \, 1-tan^{-1}\,0\right]$
$=4-4\frac{\pi}{4}=4-\pi$