If $\int \frac{d x}{(x + 1) \sqrt{2 x^{2} + 3 x + 4}}$
$= K \log f (x) + C$ , then

Moderate

A

$K = - \frac{1}{\sqrt{3}}, f (x)$
$= \frac{5 - x}{\sqrt{23} (x + 1)} + \sqrt{{(\frac{5 - x}{\sqrt{23} (x + 1)})}^{2} + \frac{23}{26}}$
B

$K = \frac{1}{\sqrt{3}} f (x) = \frac{4 - x}{(x + 1)} + \sqrt{{(\frac{4 - x}{(x + 1)})}^{2} + \frac{23}{16}}$
C

$K = 2, f (x) = y + \sqrt{y^{2} + 1, y} = \frac{5 - x}{x + 1}$
D

none of these

Solution

Putting $x + 1 = 1 / y$ , the given integral can be
written as $- \int \frac{d y}{\sqrt{3 y^{2} - y + 2}}$
$= - \frac{1}{\sqrt{3}} \int \frac{d y}{\sqrt{(y - 1 / 6)^{2} + (\sqrt{23} / 6)^{2}}}$
$\begin{array}{l} = - \frac{1}{\sqrt{3}} \log |(y - 1 / 6) + \sqrt{(y - 1 / 6)^{2} + (\sqrt{23 / 6})^{2}}| + C \\ = - \frac{1}{\sqrt{3}} \log |\frac{5 - x}{6 (x + 1)} + \sqrt{{(\frac{5 - x}{6 (x + 1)})}^{2} + {(\frac{\sqrt{23}}{6})}^{2}}| + C . \end{array}$

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