JEE 2023 — Mathematics PYQ

$\begin{array}{l}\mathrm{Put}1+\mathrm{x}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}\Rightarrow \mathrm{x}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}-1\\ \Rightarrow \left(x{e}^x+e^x\right)dx=dt\end{array}$

$\Rightarrow e^x\left(x+1\right)dx=dt$

$\therefore I=\int \frac{dt}{e^x.x{t}^2}=\int \frac{dt}{(t-1)t^2}$

Let $\frac{1}{t^2(t-1)}=\frac{\mathrm{A}}{(t-1)}+\frac{\mathrm{B}t+\mathrm{C}}{t^2}$

$\Rightarrow 1=$A$t^2+($B$t+$C$)(t-1)$

Comparing coefficients of $t^2,t$ and constant terms,

we get

A+ B= 0, C- B= 0, - C= 1

On solving above equations, we get

$\mathbf{C}=-1,=\bar{\mathbf{B}}$,A=1

$\therefore$I$=\int \frac{1}{t-1}dt+\int \frac{-t-1}{t^2}dt$

$=\int \frac{1}{t-1}dt-\int \frac{1}{t}dt-\int \frac{1}{t^2}dt$

$=\log \left|t-1\right|-\log \left|t\right|+\frac{1}{t}+$C

$\Rightarrow$I$=\log \left|x{e}^x\right|-\log \left|1+x{e}^x\right|+\frac{1}{1+r{e}^x}+c$

\begin{aligned}
 & =\log\left|\frac{xe^{x}}{1+xe^{x}}\right|+\frac{1}{1+xe^{x}}+\mathrm{C} \\
 & \mathrm{Now},\lim_{x\to\infty}\mathrm{I}(x)=0 \\
 & \Rightarrow\lim_{x\to\infty}\left\{\log\left|\frac{xe^{x}}{1+xe^{x}}\right|+\frac{1}{1+xe^{x}}+\mathrm{C}\right\}=0 \\
 & \Rightarrow\lim_{x\to\infty}\left\{\log\left(\frac{e^x}{\frac{1}{x}+e^x}\right)+\frac{\frac{1}{x}}{\frac{1}{x}+e^x}+\mathbf{C}\right\} \\
 & \Rightarrow0+0+C=0\Rightarrow C=0 \\
 & \therefore\mathrm{I}(x)=\log\left|\frac{xe^{x}}{1+xe^{x}}\right|+\frac{1}{1+xe^{x}} \\
 & \Rightarrow\mathrm{I}(1)=\log\left|\frac{e}{1+e}\right|+\frac{1}{1+e}=1-\log(1+e)+\frac{1}{1+e} \\
 & =\frac{2+e}{1+e}-\log|1+e|
\end{aligned}