If , where C is the constant of integration, then equals: [2025]

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Solution

Q.
If $\int e^{x} (\frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + \frac{\sin^{- 1} x}{({(1 - x^{2})}^{3 / 2})} + \frac{x}{1 - x^{2}}) d x = g (x) + C$ , where C is the constant of integration, then $g (\frac{1}{2})$ equals: [2025]

1 $\frac{π}{6} \sqrt{\frac{e}{2}}$

2 $\frac{π}{6} \sqrt{\frac{e}{3}}$

3 $\frac{π}{4} \sqrt{\frac{e}{2}}$

4 $\frac{π}{4} \sqrt{\frac{e}{3}}$

Ans.
(2)

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

Now, $\frac{d}{d x} (\frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}}) = \frac{\sin^{- 1} x}{{(1 - x^{2})}^{3 / 2}} + \frac{x}{1 - x^{2}}$

$\Rightarrow I = e^{x} \cdot \frac{x \sin^{- 1} x}{\sqrt{1 - x^{2}}} + C$

$[∵ \int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + c]$

= g(x) + c [Given]

$∴ g (x) = \frac{x e^{x} \sin^{- 1} x}{\sqrt{1 - x^{2}}}$

Hence, $g (1 / 2) = \frac{e^{1 / 2}}{2} \cdot \frac{\frac{π}{6} \times 2}{\sqrt{3}} = \frac{π}{6} \sqrt{\frac{e}{3}}$ .

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