Interpret the differential equation

The given equation is $(1-x^2)\,dy = \left(xy+(x^3+2)\sqrt{1-x^2}\right)dx$

So, $\frac{dy}{dx} = \frac{xy+(x^3+2)\sqrt{1-x^2}}{1-x^2}.$

Rearranging, $\frac{dy}{dx}-\frac{x}{1-x^2}y = \frac{x^3+2}{\sqrt{1-x^2}}.$

This is a linear differential equation: $\frac{dy}{dx}+P(x)y=Q(x),$ where $P(x)=-\frac{x}{1-x^2}, \qquad Q(x)=\frac{x^3+2}{\sqrt{1-x^2}}.$

Find the integrating factor

$IF = e^{\int P(x)dx} = e^{\int -\frac{x}{1-x^2}dx}.$

Let $u=1-x^2 \implies du=-2x\,dx.$ Then, $\int -\frac{x}{1-x^2}dx = \frac12\int \frac{du}{u} = \frac12\ln(1-x^2).$

Hence, $IF=e^{\frac12\ln(1-x^2)} = \sqrt{1-x^2}.$

Multiply the equation by the integrating factor

$\sqrt{1-x^2}\,\frac{dy}{dx}-\frac{x}{\sqrt{1-x^2}}y = x^3+2.$

The left side is $\frac{d}{dx}\left(y\sqrt{1-x^2}\right).$

Therefore, $\frac{d}{dx}\left(y\sqrt{1-x^2}\right)=x^3+2.$

Integrate both sides

$y\sqrt{1-x^2} = \int (x^3+2)dx = \frac{x^4}{4}+2x+C.$

So the solution is $y\sqrt{1-x^2} = \frac{x^4}{4}+2x+C.$

Use the condition hidden in the question

The printed statement appears as:

$\dots, -1$-$1 \text{ is equal to } \underline{\hspace{2cm}}$

This is clearly the standard notation for evaluating between limits $x=-1$ and $x=1$, i.e. $\left[y\sqrt{1-x^2}\right]_{-1}^{1}.$

From the solved equation, $y\sqrt{1-x^2} = \frac{x^4}{4}+2x+C.$

Hence,

Final value

$\boxed{4}$

Comparison with stored answer

The stored correct answer is $320$, but the differential equation directly gives $\left[y\sqrt{1-x^2}\right]_{-1}^{1}=4.$

So I do not agree with the stored answer. It appears the scanned/text version may have omitted the actual expression to be evaluated, and if it is indeed $\left[y\sqrt{1-x^2}\right]_{-1}^{1}$, then the correct integer should be $4$, not $320$.