Let $$y = y(x)$$ be the solution of the differential equation $$(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x$$, $$y(0) = 1$$. Then $$\int_{-3}^{3} y(x) \, dx$$ is :

The given differential equation is$$(x^{2}+1)\,y' \;-\;2x\,y \;=\;(x^{4}+2x^{2}+1)\cos x,\qquad y(0)=1$$

Step 1: Write it in standard linear form $$y'+P(x)\,y=Q(x).$$
Divide by $$(x^{2}+1):$$
$$y' \;-\;\frac{2x}{x^{2}+1}\,y \;=\;\frac{x^{4}+2x^{2}+1}{x^{2}+1}\,\cos x.$$

Notice that $$x^{4}+2x^{2}+1=(x^{2}+1)^{2},$$ so the right-hand side simplifies to$$(x^{2}+1)\cos x.$$
Thus$$y' \;-\;\frac{2x}{x^{2}+1}\,y \;=\;(x^{2}+1)\cos x\qquad -(1)$$

Step 2: Find the integrating factor (I.F.).
For $$y' + P(x)\,y = Q(x)$$, I.F. is $$e^{\int P(x)\,dx}.$$ Here $$P(x)= -\frac{2x}{x^{2}+1}.$$ Therefore$$\text{I.F.}=e^{\int -\frac{2x}{x^{2}+1}\,dx} =e^{-\ln(x^{2}+1)} =\frac{1}{x^{2}+1}.$$

Step 3: Multiply equation $$-(1)$$ by the I.F.

$$\frac{1}{x^{2}+1}\,y' \;-\;\frac{2x}{(x^{2}+1)^{2}}\;y \;=\;\cos x$$

The left side is the derivative of $$\frac{y}{x^{2}+1}$$ because $$\frac{d}{dx}\left(\frac{y}{x^{2}+1}\right)= \frac{y'(x^{2}+1)-y(2x)}{(x^{2}+1)^{2}} =\frac{1}{x^{2}+1}\,y' -\frac{2x}{(x^{2}+1)^{2}}\,y.$$

Hence$$\frac{d}{dx}\left(\frac{y}{x^{2}+1}\right)=\cos x.$$(Integrate)

Step 4: Integrate both sides.

$$\frac{y}{x^{2}+1}= \int \cos x\,dx = \sin x + C$$

Therefore$$y(x)=(x^{2}+1)\bigl(\sin x + C\bigr).$$

Step 5: Apply the initial condition $$y(0)=1.$$ At $$x=0$$,
$$1=(0^{2}+1)\bigl(\sin 0 + C\bigr)=1\cdot(0+C)=C.$$ So $$C=1.$$

Hence the required solution is$$y(x)=(x^{2}+1)\bigl(\sin x + 1\bigr).$$

Step 6: Evaluate $$\displaystyle\int_{-3}^{3}y(x)\,dx.$$
$$\int_{-3}^{3}y(x)\,dx =\int_{-3}^{3}(x^{2}+1)\bigl(\sin x + 1\bigr)\,dx$$ $$=\int_{-3}^{3}(x^{2}+1)\sin x\,dx \;+\;\int_{-3}^{3}(x^{2}+1)\,dx \qquad -(2)$$

Case 1: $$\displaystyle\int_{-3}^{3}(x^{2}+1)\sin x\,dx$$
$$(x^{2}+1)$$ is even, $$\sin x$$ is odd ⟹ their product is odd.
The integral of an odd function over $$[-a,a]$$ is $$0.$$
So this part contributes $$0.$$

Case 2: $$\displaystyle\int_{-3}^{3}(x^{2}+1)\,dx =\int_{-3}^{3}x^{2}\,dx+\int_{-3}^{3}1\,dx$$ Even function property:
$$\int_{-3}^{3}x^{2}\,dx=2\int_{0}^{3}x^{2}\,dx =2\left[\frac{x^{3}}{3}\right]_{0}^{3}=2\cdot\frac{27}{3}=18.$$ Constant term:
$$\int_{-3}^{3}1\,dx = 6.$$ Therefore the second integral equals $$18+6=24.$$

Combining both cases in $$-(2)$$:
$$\int_{-3}^{3}y(x)\,dx = 0 + 24 = 24.$$

Thus the value of $$\displaystyle\int_{-3}^{3}y(x)\,dx$$ is $$24,$$ which corresponds to Option A.