Answer
$\dfrac{1}{4} \ln (\dfrac{13}{3})$
Work Step by Step
Given: $I=\int_{2}^{3} \dfrac{x}{2x^2-5}\ dx$
Let us consider that $u=2x^2-5 \implies dx=\dfrac{du}{4x}$
Now, we have $I= \int_{2}^{3} \dfrac{x}{u} (\dfrac{du}{4x})$
or, $=\dfrac{1}{4} \int_{2}^{3} \dfrac{1}{u} \ du$
or, $=\dfrac{1}{4} [\ln |u|]_2^3 +C$
or, $=\dfrac{1}{4} [\ln |2x^2-5|]_2^3 +C$
or, $=\dfrac{1}{4} [\ln |2(3)^2-5|-\dfrac{1}{4} [\ln |2(2)^2-5|$
or, $=\dfrac{1}{4} \ln (\dfrac{13}{3})$
