The correct option is A 2 tan^ 1(√(x+1))+c nbsp;∫1/(x+2)√(x+1)dx nbsp;√(x+1)=t=1/2√(x+1)dx=dt nbsp;=dx/√(x+1)=2dt nbsp;∫2dt/(√(x+1))^2 ...

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Derivative of Standard Functions

∫1/x+2√x+1dx=

Question

$\int \frac{1}{(x + 2) \sqrt{x + 1}} d x =$

2 {tan}^{- 1} (\sqrt{x + 1}) + c

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2 {tan}^{- 1} (\sqrt{x - 1}) + c

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2 {tanh}^{- 1} (\sqrt{x + 1}) + c

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{tan}^{- 1} (\sqrt{x - 1}) + c

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Solution

The correct option is A $2 {tan}^{- 1} (\sqrt{x + 1}) + c$
$\int \frac{1}{(x + 2) \sqrt{x + 1}} d x$
$\sqrt{x + 1} = t = \frac{1}{2 \sqrt{x + 1}} d x = d t$
$= \frac{d x}{\sqrt{x + 1}} = 2 d t$
$\int \frac{2 d t}{(\sqrt{x + 1})^{2} + 1} \int \frac{2 d t}{t^{2} + 1} = 2 \int \frac{d t}{t^{2} + 1} = 2 {tan}^{- 1} (t) + c$
$2 {tan}^{- 1} (\sqrt{x + 1}) + c$

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