Question

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

• A. $3$
• B. $9$
• C. $8$
• D. $10$

Answer 1

The correct option is B

$9$

let $t=9x+3{\tan }^{-1}x$
$dt=9+\frac{3}{1+x^2}$
$dt=\frac{12+9x^2}{1+x^2}$
$\alpha =\underset{0}{\overset{\frac{9+3\pi }{4}}{\int }}{e}^{t}dx$
$\alpha =[e^t{]}_0^{\frac{9+3\pi }{4}}$ $=[e^{\frac{9+3\pi }{4}}-1]$
${\log }_e|1+\alpha |-\frac{3\pi }{4}={\log }_e|1+e^{\frac{9+3\pi }{4}}-1|-\frac{3\pi }{4}$
$=9+\frac{3\pi }{4}-\frac{3\pi }{4}=9$