_0^ , ,5 , e^x , , e^x , - ,1 e^x , + ,3 dx = - Vidyadip

MCQ

$\int\limits_0^{\ln \,\,5} {\,\frac{{{e^x}\,\,\sqrt {{e^x}\, - \,1} }}{{{e^x}\, + \,3}}} $ $dx =$

✓
$4 - \pi$
B
$6 - \pi$
C
$5 - \pi$
D
$None$

✓

Answer

Correct option: A.

$4 - \pi$

a
Let $I=\int \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3}$

Substitute $t=e^{x} \Rightarrow d t=e^{x} d x,$ we get

$I=\int \frac{\sqrt{t-1}}{t+3} d t$

Substitute $u=\sqrt{t-1} \Rightarrow d u=\frac{1}{2 \sqrt{t-1}} d t,$ we get

$I=2 \int \frac{u^{2}}{u^{2}+4} d u=2 \int\left(1-\frac{4}{u^{2}+4}\right) d u=2 \int d u-2 \int \frac{4}{u^{2}+4} d u$

$=2 u-4 \tan ^{-1}\left(\frac{u}{2}\right)=2 \sqrt{t-1}-4 \tan ^{-1}\left(\frac{\sqrt{t-1}}{2}\right)$

$=2 \sqrt{e^{x}-1}-4 \tan ^{-1}\left(\frac{\sqrt{e^{x}-1}}{2}\right)$

Therefore, $\int_{0}^{\log 5} I d x=\left[2 \sqrt{e^{x}-1}-4 \tan ^{-1}\left(\frac{\sqrt{e^{x}-1}}{2}\right)\right]_{0}^{\log 5}=4-\pi$

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