JEE Main 2025 — 7 April, Evening Shift — Question 36

$\left(1+x^{2}\right) \frac{d y}{d x}-2 x y=\left(x^{4}+2 x^{2}+1\right) \cos x$

$\frac{d y}{d x}-\left(\frac{2 x}{1+x^{2}}\right) y=\frac{\left(x^{2}+1\right) \cos x}{\left(x^{2}+1\right)}$

$\mathrm{IF}=e^{-\int \frac{2 x}{1+x^{2}} d x}=\frac{1}{1+x^{2}}$

$\frac{y}{1+x^{2}}=\int \cos x d x$ $\frac{y}{1+x^{2}}=\int \sin x+c$

$\because y(0)=1$

$\Rightarrow 1=c$

$\therefore y=(1+\sin x)\left(1+x^{2}\right)$

$\int_{-3}^{3} y(x) d x \int_{-3}^{3}(1+\sin x)\left(1+x^{2}\right) d x$

$=\int_{0}^{3} 2\left(1+x^{2}\right) d x$

$\left.=2 x+\frac{2 x^{3}}{3}\right]_{0}^{3}$

$=6+18=24$