$\begin{aligned}
& \left(x^2+1\right) \frac{d y}{d x}-2 x y=\left(x^4+2 x^2+1\right) \cos x \
& \frac{d y}{d x}-\left(\frac{2 x}{x^2+1}\right) y=\frac{\left(x^2+1\right)^2 \cos x}{\mathrm{cx}^2+1}=\left(\mathrm{x}^2+1\right) \cos x \
& \text { (Linear D.E) } \
& P=\frac{-2 x}{x^2+1}, Q=\left(x^2+1\right) \cos x \
& \text { I.F }=e^{\int P d x}=e^{\int \frac{-2 x}{x^2+1} d x}=\frac{1}{x^2+1} \
& y \cdot \frac{1}{x^2+1}=\int\left(x^2+1\right) \cos x \cdot \frac{1}{x^2+1} d x \
& \frac{y}{x^2+1}=\sin x+c \Rightarrow y \cos =1 \Rightarrow c=1 \
& y=\left(x^2+1\right)(\sin x+1) \
& \int_{-3}^3 y d x=\int_{-3}^3\left(x^2+1\right)(\sin x+1) \
& d x=\int_{-3}^3 x^2 \sin x+x^2 \sin x+1 d x \
& \Rightarrow \int_{-3}^3 x^2 \sin x d x+\int_{-3}^3 x^2 d x+\int_{-3}^3 \sin x d x+\int_{-3}^3 1 d x \
& =0+18+0+6=24
\end{aligned}$