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

We need $$f(x) = \int \frac{(2-x^2)e^x}{\sqrt{1+x}(1-x)^{3/2}}\,dx$$, given $$f(0) = 0$$.

Since $$2 - x^2 = 1 + (1-x)(1+x)$$, the integrand splits as $$e^x\left(\frac{1}{\sqrt{1+x}(1-x)^{3/2}} + \frac{\sqrt{1+x}}{\sqrt{1-x}}\right)$$.

Let $$g(x) = \sqrt{\frac{1+x}{1-x}}$$. Then $$g'(x) = \frac{1}{(1-x)^{3/2}\sqrt{1+x}}$$.

This shows the integrand is $$e^x(g(x) + g'(x))$$.

Using the identity $$\int e^x(g + g')dx = e^x g + C$$ gives $$f(x) = e^x \sqrt{\frac{1+x}{1-x}} + C$$.

Substituting $$f(0) = 0$$ yields $$1 + C = 0$$, so $$C = -1$$.

Therefore $$f\bigl(\tfrac12\bigr) = \sqrt{e}\sqrt{3} - 1 = \sqrt{3e} - 1$$, and the correct answer is Option 4: $$\sqrt{3e} - 1$$.