First slide

Methods of integration

Question

$\int \frac{dx}{(x + 2) \sqrt{x^{2} + 4 x + 8}}$ is equal to

Moderate

A

$- \frac{1}{2} \log |\frac{1}{(x + 2)} + \sqrt{\frac{1}{(x + 2)^{2}} + \frac{1}{4}}| + C$
B

$- \log |\frac{1}{(x + 2)} + \sqrt{\frac{1}{(x + 2)^{2}} + \frac{1}{4}}| + C$
C

$\frac{1}{2} \log |\frac{1}{(x + 2)} + \sqrt{\frac{1}{(x + 2)^{2}} + \frac{1}{4}}| + C$
D

None of the above

Solution

Put $x + 2 = \frac{1}{t} \Rightarrow dx = \frac{- dt}{t^{2}}$
Now $x^{2} + 4 x + 8 = (x + 2)^{2} + 4$
So $\begin{array}{l} = \int \frac{\frac{- dt}{t^{2}}}{\frac{1}{t} \sqrt{\frac{1}{t^{2}} + 4}} = - \int \frac{dt}{\sqrt{1 + 4 t^{2}}} \\ = - \frac{1}{2} \int \frac{dt}{\sqrt{t^{2} + \frac{1}{4}}} = - \frac{1}{2} \ln |t + \sqrt{t^{2} + \frac{1}{4}}| + C \\ = - \frac{1}{2} \ln |\frac{1}{x + 2} + \sqrt{\frac{1}{(x + 2)^{2}} + \frac{1}{4}}| + C \end{array}$

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