JEE Advanced 2015 Paper 2Question 1
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 $\left(\log_e|1+\alpha|-\frac{3\pi}{4}\right)$ is
Definite Integration
Answer$9$
Solution
Concept. The fraction is $9+\frac{3}{1+x^2}$, the derivative of the exponent. Formulas used. (1) $\dfrac{d}{dx}e^{u(x)}=u'(x)\,e^{u(x)}$Step 1 - Recognise. $u=9x+3\tan^{-1}x$ gives $u'=9+\dfrac{3}{1+x^2}=\dfrac{12+9x^2}{1+x^2}$. By (1), the integrand is $\dfrac{d}{dx}e^{u}$. Step 2 - Evaluate. $$\alpha=e^{9+3\pi/4}-e^0=e^{9+3\pi/4}-1.$$ Step 3 - Value. $\ln|1+\alpha|=9+\dfrac{3\pi}{4}$, so the answer is $9$. Answer: 9