JEE Main 2026 (Online) 2nd April Evening ShiftQuestion 1
Let $f(x)=\int\left(\frac{16x+24}{x^{2}+2x-15}\right)dx$. If $f(4)=14\log_e(3)$ and $f(7)=\log_e(2^{\alpha}\cdot 3^{\beta})$, $\alpha,\beta\in\mathbf{N}$, then $\alpha+\beta$ is equal to :
Indefinite Integration
Solution
Concept. Partial fractions give $f$; $f(4)$ fixes the constant. Formulas used. (1) $\int\frac{dx}{x-a}=\ln|x-a|$Step 1 - Partial fractions. $x^2+2x-15=(x+5)(x-3)$ and $$16x+24=A(x-3)+B(x+5):\quad x=3\Rightarrow B=9,\qquad x=-5\Rightarrow A=7.$$ Step 2 - Integrate. By (1), $f(x)=7\ln|x+5|+9\ln|x-3|+C$. Step 3 - Constant. $f(4)=7\ln9+0+C=14\ln3$, so $C=0$. Step 4 - $f(7)$. $$f(7)=7\ln12+9\ln4=7(2\ln2+\ln3)+18\ln2=32\ln2+7\ln3=\ln\left(2^{32}\cdot3^7\right).$$ So $\alpha+\beta=32+7=39$. Answer: (C)