Correct option is: (1) 24
\(\left(1+x^2\right) \frac{d y}{d x}-2 x y=\left(x^4+2 x^2+1\right) \cos x \)
\(\frac{d y}{d x}-\left(\frac{2 x}{1+x^2}\right) y=\frac{\left(x^2+1\right) \cos x}{\left(x^2+1\right)} \)
\( \text { IF }=e^{-\int \frac{2 x}{1+x^2} d x}=\frac{1}{1+x^2} \)
\( \frac{y}{1+x^2}=\int \cos x d x \)
\( \frac{y}{1+x^2}=\int \sin x+c \)
\( \because\ y(0)=1 \)
\( \Rightarrow 1=c \)
\( \therefore\ y=(1+\sin x)\left(1+x^2\right) \)
\(\int\limits_{-3}^3 y(x) d x \int\limits_{-3}^3(1+\sin x)\left(1+x^2\right) d x \)
\(= \int\limits_{0}^3 2(1+x^2) dx\)
\(=2x+\frac{2x^3}{3}]_0^3\)
= 6 + 18 = 24
