If $\int \frac{1}{(1 + x) \sqrt{x}} d x = f (x) + A$ , where A is any arbitary constant, then the function f(x) is

2 {tan}^{- 1} x

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2 {tan}^{- 1} \sqrt{x}

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2 {cot}^{- 1} \sqrt{x}

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l o g_{e} (1 + x)

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Solution

The correct option is B $2 {tan}^{- 1} \sqrt{x}$
Given,

$\int \frac{1}{(1 + x) \sqrt{x}} d x$

put $u = \sqrt{x}$

$= \int \frac{2}{1 + u^{2}} d u$

$= 2 {tan}^{- 1} (u)$

$= 2 {tan}^{- 1} (\sqrt{x}) + C$

$= f (x) + A$

$∴ f (x) = 2 {tan}^{- 1} (\sqrt{x})$

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x

[- 10 π, 20 π]

f (x) = s g n (g (x)),

5 a

a

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