Example 59. Evaluate \(\ds \int t^3 \ln t \ dt \).
We choose \(u\) and \(dv\) as follows:
\begin{equation*}
u = \ln t \hspace{2cm} dv = t^3 \ dt.
\end{equation*}
Then we find \(du\) by taking the derivative of \(u\) and \(v\) by taking the antiderivative of \(dv\text{:}\)
\begin{equation*}
du = \frac{1}{t}dt \hspace{2cm} v = \frac{1}{4}t^4.
\end{equation*}
Thus we have:
\begin{align*}
\int t^3 \ln t dt =\amp\ \int \ln t \cdot t^3dt \\
=\amp\ \int \ub{\ln t}_{u} \cdot \ub{t^3 dt}_{dv} \\
=\amp\ \int u\cdot dv \\
=\amp\ u\cdot v - \int v \cdot du \\
=\amp\ \ln t \cdot \frac{1}{4}t^4 - \int \frac{1}{4}t^4
\cdot \frac{1}{t}dt \\
=\amp\ \frac{1}{4}t^4\ln t - \frac{1}{4}\int t^3 dt \\
=\amp\ \frac{1}{4}t^4\ln t - \frac{1}{4}\cdot \frac{1}{4}t^4 + C \\
=\amp\ \frac{1}{4}t^4\ln t - \frac{1}{16}t^4 + C
\end{align*}
