Mathematics - Definite Integration Question with Solution | TestHub

$\alpha = \underset{0}{\overset{1}{\int }}\left(e^{9x+3{\tan }^{-1}x}\right) \left(\frac{12+9x^2}{1+x^2}\right) dx$
$= \underset{0}{\overset{1}{\int }}\left(e^{9x+3{\tan }^{-1}x}\right) \left(9+\frac{3}{1+x^2}\right) dx$
$= \left(e^{9x+3{\tan }^{-1}x}\right){|}_0^1$
$= e^{9+\frac{3\pi }{4}}-1$
$\Rightarrow \ln \mathrm{ }\left|(1+\mathrm{\alpha })\right|-\mathrm{ }\frac{3\mathrm{\pi }}{4}=\ln \left|\left(1+{\mathrm{e}}^{9+\frac{3\mathrm{\pi }}{4}}-1\right)\right|-\frac{3\mathrm{\pi }}{4}=\ln \left|{\mathrm{e}}^{9+\frac{3\mathrm{\pi }}{4}}\right|-\frac{3\mathrm{\pi }}{4}=9$ .