Let f(x) = int frac{(2-x^2)e^x}{(sqrt{1+x})(1 | vedclass

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(A) We have $f(x) = \int e^x \left( \frac{2-x^2}{\sqrt{1+x}(1-x)^{3/2}} \right) dx$. Rewrite the numerator as $(1-x^2) + 1$: $f(x) = \int e^x \left( \

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$\sqrt{3e}-1$

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(A) We have $f(x) = \int e^x \left( \frac{2-x^2}{\sqrt{1+x}(1-x)^{3/2}} \right) dx$. Rewrite the numerator as $(1-x^2) + 1$: $f(x) = \int e^x \left( \frac{1-x^2}{\sqrt{1+x}(1-x)^{3/2}} + \frac{1}{\sqrt{1+x}(1-x)^{3/2}} \right) dx$. Simplify the first term: $\frac{(1-x)(1+x)}{\sqrt{1+x}(1-x)^{3/2}} = \frac{\sqrt{1+x}}{\sqrt{1-x}} = \sqrt{\frac{1+x}{1-x}}$. Let $g(x) = \sqrt{\frac{1+x}{1-x}}$. Then $g'(x) = \frac{1}{2} \left( \frac{1+x}{1-x} \right)^{-1/2} \cdot \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{1}{2} \sqrt{\frac{1-x}{1+x}} \cdot \frac{2}{(1-x)^2} = \frac{1}{\sqrt{1+x}(1-x)^{3/2}}$. Thus, the integral is of the form $\int e^x (g(x) + g'(x)) dx = e^x g(x) + C$. So, $f(x) = e^x \sqrt{\frac{1+x}{1-x}} + C$. Given $f(0) = 0$, we have $0 = e^0 \sqrt{\frac{1}{1}} + C \implies 0 = 1 + C \implies C = -1$. Therefore, $f(x) = e^x \sqrt{\frac{1+x}{1-x}} - 1$. For $x = \frac{1}{2}$, $f(\frac{1}{2}) = e^{1/2} \sqrt{\frac{1+1/2}{1-1/2}} - 1 = \sqrt{e} \sqrt{\frac{3/2}{1/2}} - 1 = \sqrt{e} \sqrt{3} - 1 = \sqrt{3e} - 1$.

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

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