Question
Evaluate the following definite integrals : $\int_1^2 \frac{3 x}{9 x^2-1} d x$

Answer

Let $I =\int_1^2 \frac{3 x}{9 x^2-1} d x=\int_1^2 \frac{3 x}{(3 x)^2-1} d x$
Put $3 x = t$
$\therefore 3 dx = dt$
$\therefore dx =\frac{d t}{3}$
When $x=1, t=3 \times 1=3$
When $x=2, t=3 \times 2=6$
$
\begin{aligned}
\therefore I & =\int_3^6 \frac{t}{t^2-1} \cdot \frac{d t}{3}=\frac{1}{6} \int_3^6 \frac{2 t}{t^2-1} d t \\
& =\frac{1}{6}\left[\log \left|t^2-1\right|\right]_3^6 \quad \ldots\left[\because \frac{d}{d t}\left(t^2-1\right)=2 t\right] \\
& =\frac{1}{6}[\log 35-\log 8] \\
& =\frac{1}{6} \log \left(\frac{35}{8}\right) .
\end{aligned}
$
Alternative Method :
$
\begin{aligned}
& \int_1^2 \frac{3 x}{9 x^2-1} d x \\
& =\frac{1}{6} \int_1^2 \frac{18 x}{9 x^2-1} d x \\
& =\frac{1}{6}\left[\log \left|9 x^2-1\right|\right]_1^2 \\
& =\frac{1}{6}[\log 35-\log 8] \\
& =\frac{1}{6} \log \left(\frac{35}{8}\right) .
\end{aligned}
$

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