Evaluate: $\;$ $\displaystyle \int_{2}^{3} \dfrac{x^3 + 1}{x \left(x - 1\right)} \; dx$
Let $\;$ $I = \displaystyle \int_{2}^{3} \dfrac{x^3 + 1}{x \left(x - 1\right)} \; dx$
$\begin{aligned}
i.e. \; I & = \int_{2}^{3} \dfrac{\left(x - 1\right) \left(x^2 + x + 1\right)}{x \left(x - 1\right)} \; dx \;\;\; \left[\text{Note: } a^3 - b^3 = \left(a - b\right) \left(a^2 + ab + b^2\right)\right] \\\\
& = \int_{2}^{3} \dfrac{x^2 + x + 1}{x} \; dx \\\\
& = \int_{2}^{3} x \; dx + \int_{2}^{3} dx + \int_{2}^{3} \dfrac{1}{x} \; dx \\\\
& = \left[\dfrac{x^2}{2}\right]_{2}^{3} + \left[x\right]_{2}^{3} + \left[\log x\right]_{2}^{3} \\\\
& = \dfrac{1}{2} \left(3^2 - 2^2\right) + \left(3 - 2\right) + \log 3 - \log 2 \\\\
& = \dfrac{7}{2} + \log \left(\dfrac{3}{2}\right)
\end{aligned}$