JEE Main 2024 — 5 April, Shift 2 — Question 22

$\quad \mathrm{IF}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{2}} \mathrm{dx}}=\mathrm{e}^{\frac{-1}{1+\mathrm{x}^{2}}}$

$\mathrm{y} \cdot \mathrm{e}^{\frac{-1}{1+\mathrm{x}^{2}}}=\int \mathrm{x} \cdot \mathrm{e}^{\frac{1}{1+\mathrm{x}^{2}}} \cdot \mathrm{e}^{\frac{-1}{1+\mathrm{x}^{2}} \mathrm{dx}}$

$y \cdot e^{\frac{-1}{1+x^{2}}}=\frac{x^{2}}{2}+c$

$(0,0) \Rightarrow \mathrm{C}=0$

$y(x)=\frac{x^{2}}{2} e^{\frac{1}{1+x^{2}}}$

$f(x)=\frac{x^{2}}{2}$

$A=\int_{-2}^{4}(x+4)-\frac{x^{2}}{2} d x=18$