QJEEAdv15Ma4e4a: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Ma4e4a Solution: If where takes only principal values, then the value of is

$\alpha = \int\limits_0^1 {{e^{(9x + 3{{\tan }^{ - 1}}x)}}\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)dx}$

Set $9x + 3{\tan ^{ - 1}}x = t$

so that ${{dt} \over {dx}} = 9 + {3 \over {1 + {x^2}}} = {{12 + 9{x^2}} \over {1 + {x^2}}}$

We have, $\alpha = \int\limits_0^{9 + {{3\pi } \over 4}} {{e^t}dt = {e^{9 + {{3\pi } \over 4}}} - 1}$

$\therefore$ $\ln \left| {\alpha + 1} \right| = 9 + {{3\pi } \over 4}$

Thus $\ln \left| {\alpha + 1} \right| - {{3\pi } \over 4} = 9$