If $\alpha=\int\limits_0^1 e^{9 x+3 \tan ^{-1} x}\left(\frac{12+9 x^2}{1+x^2}\right) d x$, where $\tan ^{-1} x$ takes only principal values, then the value of $\left(\log _e|1+\alpha|-\frac{3 \pi}{4}\right)$, is

9

We have

$\alpha=\int\limits_0^1 e^{9 x+3 \tan ^{-1} x}\left(\frac{12+9 x^2}{1+x^2}\right) d x$

$\Rightarrow \alpha=\int\limits_0^{9+\frac{3 \pi}{4}} e^t d t$, where $t=9 x+3 \tan ^{-1} x$

$\Rightarrow \alpha=e^{9+\frac{3 \pi}{4}}-1$

$\Rightarrow 1+\alpha=e^{9+\frac{3 \pi}{4}}$

$\Rightarrow \log _e|1+\alpha|=9+\frac{3 \pi}{4} \Rightarrow \log _e|1+\alpha|-\frac{3 \pi}{4}=9$