Let $f(x)=\int \frac{\left(2-x^2\right) \cdot \mathrm{e}^x}{(\sqrt{1+x})(1-x)^{3 / 2}} \mathrm{~d} x$. If $f(0)=0$, then $f\left(\frac{1}{2}\right)$ is equal to:

Question

Accepted Answer

C. 1. **Why**: Combine the square-root factors in the denominator into a single expression, since $\sqrt{1+x}(1-x)=\sqrt{1-x^2}\cdot(1-x)$... actually simplify directly using $\sqrt{1+x}\cdot(1-x)^{3/2}=\sqrt{1-x^2}\,(1-x)$.
$$
f(x)=\int\frac{(2-x^2)e^x}{\sqrt{1-x^2}\,(1-x)}\,dx
$$

2. **Why**: Split the numerator $2-x^2$ as $1+(1-x^2)$ so the integral separates into two recognizable pieces.
$$
f(x)=\int e^x\left[\frac{1}{\sqrt{1-x^2}(1-x)}+\frac{1+x}{\sqrt{1-x^2}}\right]dx
$$

3. **Why**: This matches the standard form $\int e^x[g(x)+g'(x)]dx=e^xg(x)+C$; guess $g(x)=\frac{1+x}{\sqrt{1-x^2}}$ and verify by differentiating.
$$
g'(x)=\frac{\sqrt{1-x^2}+\frac{x^2+x}{\sqrt{1-x^2}}}{1-x^2}=\frac{1}{\sqrt{1-x^2}(1-x)}
$$
This matches the first bracket term, confirming the guess.

4. **Why**: Apply the standard-form result directly.
$$
f(x)=\frac{e^x(1+x)}{\sqrt{1-x^2}}+C
$$

5. **Why**: Use the given condition $f(0)=0$ to solve for $C$.
$$
0=\frac{e^0(1)}{\sqrt1}+C \implies C=-1
$$

6. **Why**: Evaluate $f\left(\frac12\right)$ using the completed formula.
$$
f\left(\frac12\right)=\frac{e^{1/2}\cdot\frac32}{\sqrt{3}/2}-1=\sqrt{3e}-1
$$

Hence, the answer is **Option C**.

Let f(x)=∫(2−x2)⋅ex(1+x)(1−x)3/2 dxf(x)=\int \frac{\left(2-x^2\right) \cdot \mathrm{e}^x}{(\sqrt{1+x})(1-x)^{3 / 2}} \mathrm{~d} xf(x)=∫(1+x​)(1−x)3/2(2−x2)⋅ex​ dx. If f(0)=0f(0)=0f(0)=0, then f(12)f\left(\frac{1}{2}\right)f(21​) is equal to:

Let $f(x)=\int \frac{\left(2-x^2\right) \cdot \mathrm{e}^x}{(\sqrt{1+x})(1-x)^{3 / 2}} \mathrm{~d} x$ . If $f(0)=0$ , then $f\left(\frac{1}{2}\right)$ is equal to: