A) 6
B) 7
C) 8
D) 9
Correct Answer: D
Solution :
We have, |
\[\alpha =\int\limits_{0}^{1}{({{e}^{9x+3{{\tan }^{-1}}x}})\left( \frac{12+9{{x}^{2}}}{1+{{x}^{2}}} \right)dx}\] |
\[\alpha =\int\limits_{0}^{1}{({{e}^{9x+3{{\tan }^{-1}}x}})\left( 9+\frac{3}{1+{{x}^{2}}} \right)dx}\] |
Put \[9x+3{{\tan }^{-\,1}}x=t\] |
\[\Rightarrow \] \[\left( 9+\frac{3}{1+{{x}^{2}}} \right)dx=dt\] |
\[\therefore \]\[\alpha =\int\limits_{0}^{9+\frac{3\pi }{4}}{{{e}^{t}}dt=[{{e}^{t}}]_{0}^{9+\frac{3\pi }{4}}}\] |
\[\Rightarrow \]\[\alpha ={{e}^{9+\frac{3\pi }{4}}}-1\]\[\Rightarrow \]\[(\alpha +1)={{e}^{9+\frac{3\pi }{4}}}\]\[\Rightarrow \]\[{{\log }_{e}}(\alpha +1)-\frac{3\pi }{4}=9\] |
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