If $\alpha=\int_0^1\left(e^{9 x+3 \tan ^{-1} x}\right)\left(\frac{12+9 x^2}{1+x^2}\right) d x$ where $\tan ^{-1} x$ takes only principal values, then the value of $\left(\log _e|1+\alpha|-\frac{3 \pi}{4}\right)$ is
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