Key Concepts and Formulas

• Integration by parts: $\int u \, dv = uv - \int v \, du$
• The derivative of $\sin^{-1}(x)$: $\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}$
• The integral $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$

Step-by-Step Solution

Step 1: Identify the pattern We are given the integral $\int \mathrm{e}^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) \mathrm{d} x$ We want to express the integrand in the form $e^x (f(x) + f'(x))$. Let's consider $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.

Step 2: Differentiate the potential f(x) We calculate the derivative of $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$ using the quotient rule and product rule: $f'(x) = \frac{(\sqrt{1-x^2})(\sin^{-1} x + \frac{x}{\sqrt{1-x^2}}) - (x \sin^{-1} x)(\frac{-2x}{2\sqrt{1-x^2}})}{1-x^2}$ $f'(x) = \frac{\sqrt{1-x^2} \sin^{-1} x + x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}}}{1-x^2}$ $f'(x) = \frac{(1-x^2) \sin^{-1} x + x\sqrt{1-x^2} + x^2 \sin^{-1} x}{(1-x^2)^{3/2}}$ $f'(x) = \frac{\sin^{-1} x - x^2 \sin^{-1} x + x\sqrt{1-x^2} + x^2 \sin^{-1} x}{(1-x^2)^{3/2}}$ $f'(x) = \frac{\sin^{-1} x + x\sqrt{1-x^2}}{(1-x^2)^{3/2}}$ $f'(x) = \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}$

Step 3: Rewrite the integral Now we can rewrite the original integral as $\int e^x \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2} \right) dx = \int e^x (f(x) + f'(x)) dx$ where $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$ and $f'(x) = \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}$.

Step 4: Apply the formula Using the formula $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$, we have $\int e^x (f(x) + f'(x)) dx = e^x \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + C$ Thus, $g(x) = e^x \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.

Step 5: Evaluate g(1/2) We are asked to find $g\left(\frac{1}{2}\right)$. Plugging in $x = \frac{1}{2}$, we get $g\left(\frac{1}{2}\right) = e^{1/2} \frac{\frac{1}{2} \sin^{-1} \left(\frac{1}{2}\right)}{\sqrt{1-\left(\frac{1}{2}\right)^2}} = e^{1/2} \frac{\frac{1}{2} \cdot \frac{\pi}{6}}{\sqrt{1-\frac{1}{4}}} = e^{1/2} \frac{\frac{\pi}{12}}{\sqrt{\frac{3}{4}}} = e^{1/2} \frac{\frac{\pi}{12}}{\frac{\sqrt{3}}{2}} = e^{1/2} \frac{\pi}{12} \cdot \frac{2}{\sqrt{3}} = e^{1/2} \frac{\pi}{6\sqrt{3}} = \frac{\pi}{6} \sqrt{\frac{e}{3}}$

Common Mistakes & Tips

• Carefully apply the quotient and product rules when differentiating $f(x)$.
• Recognize the pattern $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
• Double-check the evaluation of $g(1/2)$.

Summary

We identified the pattern $\int e^x (f(x) + f'(x)) dx$ in the given integral by recognizing that the derivative of $\frac{x \sin^{-1} x}{\sqrt{1-x^2}}$ equals the sum of the other two terms in the integrand. Then, we used the formula $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$ to find $g(x)$. Finally, we evaluated $g(1/2)$ to get the answer.

Final Answer

The final answer is $\boxed{\frac{\pi}{6} \sqrt{\frac{\mathrm{e}}{3}}}$, which corresponds to option (A).