First slide

Methods of integration

Question

$\int \frac{1}{x (2 + 3 \log x)} d x =$

Easy

A

$\frac{1}{3} \log | 2 + 3 \log x | + c$
B

$\frac{1}{3} \log | 2 - 3 \log x | + c$
C

$\frac{1}{3} \log | 3 + 2 \cos x | + c$
D

$\frac{1}{2} \log | 3 + 3 \log x | + c$

Solution

$2 + 3 \log x = t \Rightarrow \frac{3}{2} d x = d t$
$\frac{1}{3} \int \frac{1}{t} d t = \frac{1}{3} \log | t | + c$

$= \frac{1}{3} \log | 2 + 3 \log x | + c$

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