$\int \frac{x d x}{(1 + x)^{3 / 2}}$ is equal to.

2 \frac{(2 + x)}{\sqrt{1 + x}} + C

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\frac{(2 + x)}{\sqrt{1 + x}} + C

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\frac{3}{2} \frac{3}{\sqrt{1 + x}} + C

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\frac{3}{2} \frac{(2 + x)}{\sqrt{1 + x}} + C

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Solution

The correct option is A $2 \frac{(2 + x)}{\sqrt{1 + x}} + C$
$\int \frac{x d x}{(1 + x)^{3 / 2}}$
$= \int \frac{(1 + x - 1)}{(1 + x)^{3 / 2}} d x$
$= \int \frac{1}{\sqrt{1 + x}} d x - \int \frac{1}{(1 + x)^{3 / 2}} d x$
$= 2 \sqrt{1 + x} + \frac{2}{\sqrt{1 + x}} + C$
$= \frac{2 (1 + x) + 2}{\sqrt{1 + x}}$
$= \frac{2 (2 + x)}{\sqrt{1 + x}} + C$

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