JEE Main 2023 (Online) 8th April Morning ShiftQuestion 1
Let $I(x)=\int\frac{(x+1)}{x\left(1+xe^{x}\right)^{2}}dx$, $x>0$. If $\lim_{x\to\infty}I(x)=0$, then $I(1)$ is equal to
Indefinite Integration
Solution
Concept. $(x+1)e^x$ is the derivative of $xe^x$; with $u=xe^x$ the integrand becomes rational in $u$. Formulas used. (1) $u=xe^x$: $du=(x+1)e^x\,dx$(2) $\dfrac1{u(1+u)^2}=\dfrac1u-\dfrac1{1+u}-\dfrac1{(1+u)^2}$Step 1 - Substitute. Multiply top and bottom by $e^x$. By (1), $$\frac{(x+1)\,dx}{x(1+xe^x)^2}=\frac{(x+1)e^x\,dx}{xe^x(1+xe^x)^2}=\frac{du}{u(1+u)^2}.$$ Step 2 - Integrate. By (2), $$I=\ln\frac{u}{1+u}+\frac1{1+u}+c.$$ Step 3 - Constant. As $x\to\infty$, $u\to\infty$: $\ln\frac u{1+u}\to0$ and $\frac1{1+u}\to0$. So $c=0$. Step 4 - $I(1)$. $u=e$: $$I(1)=\ln\frac{e}{e+1}+\frac1{e+1}=1+\frac1{e+1}-\ln(e+1)=\frac{e+2}{e+1}-\log_e(e+1).$$ Answer: (A)