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 principal values, then the value of $({\log }_e|1+\alpha |-\frac{3\pi }{4})$ is

Answer 1

$\alpha ={\int }_0^1\left(e^{9x+3{\tan }^{1}x}\right)\left(\frac{12+9x^2}{1+x^2}\right)dx$
$={\int }_0^1\left(e^{9x+3{\tan }^{1}x}\right)(9+\frac{3}{1+x^2})dx$
$={\left(e^{9x+3{\tan }^{1}x}\right)|}_0^1$
$=e^{9+\frac{3\pi }{4}}-1$
$\Rightarrow \ln |(1+\alpha )|-\frac{3\pi }{4}=9$