Answer
$\dfrac{1}{3} \ln (\dfrac{26}{7})$
Work Step by Step
Given: $I=\int_{2}^{3} \dfrac{x^2}{x^3-1}\ dx$
Let us consider that $u=x^3-1 \implies dx=\dfrac{du}{3x^2}$
Now, we have $I= \int_{2}^{3} \dfrac{x^2}{u} (\dfrac{du}{3x^2})$
or, $=\dfrac{1}{3} \int_{2}^{3} \dfrac{1}{u} \ du$
or, $=\dfrac{1}{3} [\ln |u|]_2^3 +C$
or, $=\dfrac{1}{3} [\ln |(x^3-1)|]_2^3 +C$
or, $=\dfrac{1}{3} [\ln |(3x^3-1)|]-\dfrac{1}{3} [\ln |(2^3-1)|]$
or, $=\dfrac{1}{3} \ln (\dfrac{26}{7})$
