Question

${\displaystyle {\int }_0^{\log 5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\cdot dx}$ = ______.

Options

• 3 + 2π

• 2 + π

• 4 – π

• 4 + π

Answer 1

${\displaystyle {\int }_0^{\log 5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\cdot dx}$ = 4 – π.

Explanation:

Putting e^x – 1 = t² in the given integral, we have

also ${\displaystyle \frac{d}{dx}(e^x-1)=\frac{d}{dx}{t}^2}$

⇒ ${\displaystyle e^x=2t\frac{dt}{dx}}$

${\displaystyle {\int }_0^{\log 5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx}$

= ${\displaystyle 2{\int }_0^2\frac{t^2}{t^2+4}dt}$

= ${\displaystyle 2{\int }_0^2\frac{t^2+4-4}{t^2+4}dt}$

= ${\displaystyle 2({\int }_0^{2}1dt-4{\int }_0^2\frac{dt}{t^2+4}}$

= ${\displaystyle 2\left[{(t-2{\tan }^{-1}\left(\frac{t}{2}\right))}_0^2\right]}$

= ${\displaystyle 2\left[(2-2\times \frac{\pi }{4})\right]}$

= 4 – π