Evaluate:
$\int \frac{1}{2 x + 3} d x$

log \sqrt{2 x + 3}

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log \sqrt{2 x - 3}

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log \sqrt{4 x + 6}

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None of the Above

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Solution

The correct option is A $log \sqrt{2 x + 3}$

$\int \frac{1}{2 x + 3} d x$

Let $t = 2 x + 3 ⟹ d t = 2 d x$ ........(differentiating with respect to $x$ )

Replacing we get,

$\int \frac{1}{2 t} d t$

$= \frac{1}{2} log t$

$= log t^{\frac{1}{2}}$

$= log \sqrt{2 x + 3} + C$

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