Question
Evalute : $\int \frac{1}{2 x+3 x \log x} d x$

Answer

$
\text { Let } \begin{aligned}
I & =\int \frac{1}{2 x+3 x \log x} d x \\
& =\int \frac{1}{(2+3 \log x)} \cdot \frac{1}{x} d x
\end{aligned}
$
Put $2+3 \log x=t \quad \therefore \frac{3}{x} d x=d t$
$\therefore \frac{1}{x} d x=\frac{d t}{3}$
$
\begin{aligned}
\therefore I & =\int \frac{1}{t} \cdot \frac{d t}{3}=\frac{1}{3} \int \frac{1}{t} d t \\
& =\frac{1}{3} \log |t|+c \\
& =\frac{1}{3} \log |2+3 \log x|+c .
\end{aligned}
$

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