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