Question

If $\alpha ={\int }_0^1\left(e^{(9x+3{\tan }^{-1}x)}\right)\left(\frac{12+9x^2}{1+x^2}\right)dx$ , where ${\tan }^{-1}x$ takes only principal values, then the value of $({\log }_e|1+\alpha |-\frac{3\pi }{4})$ is

Answer 1

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