Question
$\int_2^3 \frac{x}{x^2-1} d x$

Answer

Let $I =\int_2^3 \frac{x}{x^2-1} \cdot d x$
Put $x ^2-1= t$
$
\begin{aligned}
& \therefore 2 x \cdot dx = dt \\
& \therefore x \cdot dx =\frac{1}{2} \cdot d t
\end{aligned}
$
When $x=2, t=2^2-1=3$
When $x=3, t=3^2-1=8$
$
\begin{aligned}
& \therefore I =\int_3^8 \frac{1}{ t } \cdot \frac{ dt }{2} \\
& =\frac{1}{2} \int_3^8 \frac{ dt }{ t } \\
& =\frac{1}{2}[\log | t |]_3^8 \\
& =\frac{1}{2}(\log 8-\log 3) \\
& \therefore I =\frac{1}{2} \log \left(\frac{8}{3}\right) .
\end{aligned}
$

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