Let f ( x ) = ( 2 - x 2 ) e x ( 1 + x ) ? ( 1 - x ) 3 / 2 d x . If f ( 0 ) = 0 , then f ( 1 2 ) is equal to: [2026]

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Solution

Q.
Let $f (x)$ $= \int \frac{(2 - x^{2}) . e^{x}}{(\sqrt{1 + x}) {(1 - x)}^{3 / 2}} d x .$ If $f (0) = 0$ , then $f (\frac{1}{2})$ is equal to: [2026]

1 $\sqrt{3 e} + 1$

2 $\sqrt{2 e} - 1$

3 $\sqrt{2 e} + 1$

4 $\sqrt{3 e} - 1$

Ans.
(4)

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

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

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

$= e^{x} \sqrt{\frac{1 + x}{1 - x}} + C$

$f (x) = e^{x} \sqrt{\frac{1 + x}{1 - x}} - 1$

$f (\frac{1}{2}) = \sqrt{3 e} - 1$

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