1. We need to evaluate

$$\int e^x\left(\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{\sin^{-1}x}{(1-x^2)^{3/2}}+\frac{x}{1-x^2}\right)dx=g(x)+C.$$

We will try to recognize the integrand as the derivative of a product involving $e^x$.

2. Let

$$f(x)=\frac{\sin^{-1}x}{\sqrt{1-x^2}}.$$

Then

$$\frac{d}{dx}\big(e^x f(x)\big)=e^x\big(f(x)+f'(x)\big).$$

So we compute $f'(x)$.

3. Differentiate

$$f(x)=\sin^{-1}x\,(1-x^2)^{-1/2}.$$

Using product rule,

$$f'(x)=\frac{1}{\sqrt{1-x^2}}\cdot (1-x^2)^{-1/2}+\sin^{-1}x\cdot \frac{d}{dx}(1-x^2)^{-1/2}.$$

Now,

$$\frac{1}{\sqrt{1-x^2}}\cdot (1-x^2)^{-1/2}=\frac{1}{1-x^2},$$

and

$$\frac{d}{dx}(1-x^2)^{-1/2}=\frac{x}{(1-x^2)^{3/2}}.$$

Hence

$$f'(x)=\frac{1}{1-x^2}+\frac{x\sin^{-1}x}{(1-x^2)^{3/2}}.$$

4. Therefore,

$$f(x)+f'(x)=\frac{\sin^{-1}x}{\sqrt{1-x^2}}+\frac{1}{1-x^2}+\frac{x\sin^{-1}x}{(1-x^2)^{3/2}}.$$

This is not exactly in the given form. So let us instead try

$$f(x)=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}.$$

Then

$$\frac{d}{dx}\big(e^x f(x)\big)=e^x(f+f').$$

Now compute $f'(x)$.

5. Let

$$f(x)=x\sin^{-1}x\,(1-x^2)^{-1/2}.$$

Differentiate using product rule:

$$f'(x)=\sin^{-1}x(1-x^2)^{-1/2}+x\cdot \frac{1}{\sqrt{1-x^2}}\cdot (1-x^2)^{-1/2}+x\sin^{-1}x\cdot \frac{x}{(1-x^2)^{3/2}}.$$

Simplify:

$$f'(x)=\frac{\sin^{-1}x}{\sqrt{1-x^2}}+\frac{x}{1-x^2}+\frac{x^2\sin^{-1}x}{(1-x^2)^{3/2}}.$$

Thus

f+f'=rac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{\sin^{-1}x}{\sqrt{1-x^2}}+\frac{x}{1-x^2}+\frac{x^2\sin^{-1}x}{(1-x^2)^{3/2}}.

Combine the two terms containing $\sin^{-1}x$:

$$\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{x^2\sin^{-1}x}{(1-x^2)^{3/2}} =\frac{x\sin^{-1}x(1-x^2)+x^2\sin^{-1}x}{(1-x^2)^{3/2}} =\frac{x\sin^{-1}x}{(1-x^2)^{3/2}}.$$

So

f+f'=rac{x\sin^{-1}x}{(1-x^2)^{3/2}}+\frac{\sin^{-1}x}{\sqrt{1-x^2}}+\frac{x}{1-x^2}.

Still not matching the given integrand directly.

6. Now observe the given first two terms:

$$\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{\sin^{-1}x}{(1-x^2)^{3/2}}.$$

Factor $\sin^{-1}x$:

$$\sin^{-1}x\left(\frac{x}{\sqrt{1-x^2}}+\frac{1}{(1-x^2)^{3/2}}\right) =\sin^{-1}x\cdot \frac{x(1-x^2)+1}{(1-x^2)^{3/2}}.$$

This does not suggest immediate simplification, so let us instead check derivative of

$$F(x)=e^x\cdot \frac{\sin^{-1}x}{\sqrt{1-x^2}}.$$

From step 3,

$$F'(x)=e^x\left(\frac{\sin^{-1}x}{\sqrt{1-x^2}}+\frac{1}{1-x^2}+\frac{x\sin^{-1}x}{(1-x^2)^{3/2}}\right).$$

Now compare this with the given integrand. At this stage, it is natural to suspect a typographical mismatch in the first term of the question, because the standard exact derivative structure is

$$e^x\left(\frac{x\sin^{-1}x}{(1-x^2)^{3/2}}+\frac{\sin^{-1}x}{\sqrt{1-x^2}}+\frac{1}{1-x^2}\right),$$

which integrates to $e^x\dfrac{\sin^{-1}x}{\sqrt{1-x^2}}$.

7. Since the options involve only a clean value with $\sin^{-1}(1/2)=\pi/6$ and $\sqrt{1-(1/2)^2}=\sqrt{3}/2$, let us evaluate

$$g\left(\frac12\right)=e^{1/2}\cdot \frac{\sin^{-1}(1/2)}{\sqrt{1-1/4}} =\sqrt e\cdot \frac{\pi/6}{\sqrt{3}/2}.$$

Simplify:

$$\frac{\pi}{6}\cdot \sqrt e\cdot \frac{2}{\sqrt3} =\frac{\pi\sqrt e}{3\sqrt3} =\frac{\pi}{3}\sqrt{\frac e3}.$$

This value is not among the options.

8. Let us instead test the expression

$$g(x)=e^{x/2}\sin^{-1}x,$$

whose value at $x=1/2$ is

$$g\left(\frac12\right)=e^{1/4}\cdot \frac{\pi}{6}=\frac{\pi}{6}\sqrt[4]{e},$$

not among the options.

9. The stored answer is option A:

$$\frac{\pi}{6}\sqrt{\frac e3}.$$

That equals

$$\frac{\pi\sqrt e}{6\sqrt3}.$$

This would arise from

$$\sqrt e\cdot \frac{\pi}{6\sqrt3},$$

which is exactly half of the natural value from $e^{1/2}\dfrac{\sin^{-1}x}{\sqrt{1-x^2}}$ at $x=1/2$.

10. Therefore, the printed integrand and the options are inconsistent as written. Using the standard derivative pattern, the antiderivative should be

$$g(x)=e^x\frac{\sin^{-1}x}{\sqrt{1-x^2}},$$

which gives

$$g\left(\frac12\right)=\frac{\pi}{3}\sqrt{\frac e3}.$$

This does not match any option, including the stored answer.