Question

If $\alpha =\underset{0}{\overset{1}{\int }}\left(e^{9x+3{\tan }^{-1}x}\right)\left(\frac{12+9x^2}{1+x^2}\right)dx$, where ${\tan }^{-1}x$ takes only principle values, then the value of $(\ln |1+\alpha |-\frac{3\pi }{4})$ is

Answer 1

$\alpha =\underset{0}{\overset{1}{\int }}\left(e^{9x+3{\tan }^{-1}x}\right)\left(\frac{12+9x^2}{1+x^2}\right)dx$
Let $9x+3{\tan }^{-1}x=t$
$\left(\frac{12+9x^2}{1+x^2}\right)dx=dt$
$\Rightarrow \alpha =\underset{0}{\overset{9+\frac{3\pi }{4}}{\int }}{e}^{t}dt=e^{9+\frac{3\pi }{4}}-1$
$\Rightarrow \ln |1+\alpha |-\frac{3\pi }{4}=9$