Question
$\int_0^1 \frac{1}{2 x-3} d x$

Answer

$
\begin{aligned}
& \int_0^1 \frac{1}{2 x-3} d x=\left[\frac{\log |2 x-3|}{2}\right]_0^1 \\
& =\frac{1}{2}[\log |-1|-\log |-3|] \\
& =\frac{1}{2}(\log 1-\log 3) \\
& =-\frac{1}{2} \log 3 \text {. } \\
& \ldots[\because \log 1=0] \\
\end{aligned}
$

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