MCQ
$\int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} \,dx = $
- A$3 + 2\pi $
- ✓$4 - \pi $
- C$2 + \pi $
- Dએકપણ નહીં.
$\Rightarrow {e^x}dx = 2t\,dt$
Also as $x = 0$ to $\log 5,t = 0$ to $2$
Therefore, $\int_0^{\log 5} {\frac{{{e^x}\sqrt {{e^x} - 1} }}{{{e^x} + 3}}} dx = \int_0^2 {\frac{{2{t^2}}}{{{t^2} + 4}}dt} $
$ = 2\left[ {\int_0^2 {1dt - 4\int_0^2 {\frac{{dt}}{{{t^2} + 4}}} } } \right] = 4 - \pi $.
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