Indefinite Integration 1 Question 89
90. If $\alpha=\int _0^{1}\left(e^{9 x+3 \tan ^{-1} x}\right) \frac{12+9 x^{2}}{1+x^{2}} d x$,
where $\tan ^{-1} x$ takes only principal values, then the value of $\log _e|1+\alpha|-\frac{3 \pi}{4}$ is
(2015 Adv.)
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Answer:
Correct Answer: 90. (9)
Solution:
- Here, $\alpha=\int _0^{1} e^{\left(9 x+3 \tan ^{-1} x\right)} \frac{12+9 x^{2}}{1+x^{2}} d x$
Put $\quad 9 x+3 \tan ^{-1} x=t$
$\Rightarrow \quad 9+\frac{3}{1+x^{2}} d x=d t$
$\therefore \quad \alpha=\int _0^{9+3 \pi / 4} e^{t} d t=\left[e^{t}\right] _0^{9+3 \pi / 4}=e^{9+3 \pi / 4}-1$
$\Rightarrow \quad \log _e|1+\alpha|=9+\frac{3 \pi}{4}$
$\Rightarrow \quad \log _e|\alpha+1|-\frac{3 \pi}{4}=9$
