Evaluate $\displaystyle \int \left(1 - x^2\right) \; \log x \; dx$ $\;\;\;$ $x > 0$
$\begin{aligned}
\text{Let } I & = \int \left(1 - x^2\right) \; \log x \; dx \\\\
& = \int \log x \; dx - \int x^2 \; \log x \; dx \\\\
& = I_1 - I_2 \;\;\; \cdots (1)
\end{aligned}$
$\left[\text{Note: } \displaystyle \int u \cdot v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \right]$
Consider $I_1 = \displaystyle \int \log x \; dx$ $\;\;\; \cdots$ (2)
Here $u = \log x$ and $v = 1$
$\begin{aligned}
\therefore \; I_1 & = \log x \int dx - \int \left\{\int dx \times \dfrac{d}{dx} \left(\log x\right) \right\} \; dx \\\\
& = x \; \log x - \int x \times \dfrac{1}{x} \; dx \\\\
& = x \; \log x - x + c_1 \;\;\; \cdots (2a)
\end{aligned}$
Consider $I_2 = \displaystyle \int x^2 \; \log x \; dx$ $\;\;\; \cdots$ (3)
Here $u = \log x$ and $v = x^2$
$\begin{aligned}
\therefore \; I_2 & = \log x \int x^2 \; dx - \int \left\{\int x^2 \; dx \times \dfrac{d}{dx} \left(\log x\right) \right\} \; dx \\\\
& = \log x \times \dfrac{x^3}{3} - \int \dfrac{x^3}{3} \times \dfrac{1}{x} \; dx \\\\
& = \dfrac{x^3 \; \log x}{3} - \dfrac{1}{3} \int x^2 \; dx \\\\
& = \dfrac{x^3 \; \log x}{3} - \dfrac{x^3}{9} + c_2 \;\;\; \cdots (3a)
\end{aligned}$
$\therefore$ $\;$ In view of equations (2a) and (3a), equation (1) becomes
$I = x \; \log x - x - \dfrac{x^3 \; \log x}{x^3} + \dfrac{x^3}{9} + c $
where $c = c_1 - c_2$