If ∫dx(x+1)2x2+3x+4=Klog⁡f(x)+C, then

Putting x+1=1/y, the given integral can be written as −∫dy3y2−y+2=−13∫dy(y−1/6)2+(23/6)2=−13log⁡(y−1/6)+(y−1/6)2+(23/6)2+C=−13log⁡5−x6(x+1)+5−x6(x+1)2+2362+C.

If ∫dx(x+1)2x2+3x+4=Klog⁡f(x)+C, then

Putting x+1=1/y, the given integral can be written as −∫dy3y2−y+2=−13∫dy(y−1/6)2+(23/6)2=−13log⁡(y−1/6)+(y−1/6)2+(23/6)2+C=−13log⁡5−x6(x+1)+5−x6(x+1)2+2362+C.

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

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$K = - \frac{1}{\sqrt{3}}, f (x)$

$= \frac{5 - x}{\sqrt{23} (x + 1)} + \sqrt{{(\frac{5 - x}{\sqrt{23} (x + 1)})}^{2} + \frac{23}{26}}$

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

none of these

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

answer is D.

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Detailed Solution

Putting $x + 1 = 1 / y$ , the given integral can be

written as $- \int \frac{d y}{\sqrt{3 y^{2} - y + 2}}$

$= - \frac{1}{\sqrt{3}} \int \frac{d y}{\sqrt{(y - 1 / 6)^{2} + (\sqrt{23} / 6)^{2}}}$

$\begin{array}{l} = - \frac{1}{\sqrt{3}} \log |(y - 1 / 6) + \sqrt{(y - 1 / 6)^{2} + (\sqrt{23 / 6})^{2}}| + C \\ = - \frac{1}{\sqrt{3}} \log |\frac{5 - x}{6 (x + 1)} + \sqrt{{(\frac{5 - x}{6 (x + 1)})}^{2} + {(\frac{\sqrt{23}}{6})}^{2}}| + C . \end{array}$