\(\int (2x + 3) \sin(4x - 1)\:dx\)
\(\int u \: dv = u \: v - \int v \: du\)
\begin{equation*} \begin{split} u & = 2x + 3 \\\\ du & = 2 \: dx \end{split} \end{equation*}
\begin{equation*} \begin{split} dv & = \sin (4x - 1)\:dx \\\\ \int dv & = \int \sin (4x - 1)\:dx \\\\ v & = \int \sin (4x - 1) \: \frac{d(4x - 1)}{4} \\\\ v & = \tfrac 14 \int \sin (4x - 1) \: d(4x - 1) \\\\ v & = - \tfrac 14 \cos (4x - 1) \end{split} \end{equation*}
\begin{equation*}
\begin{split}
& \int u \: dv = u \:.\: v - \int v \: du \\\\
& \int (2x + 3) \sin(4x - 1)\:dx \\\\
& (2x + 3) \:.\: - \tfrac 14 \cos(4x - 1) - \int - \tfrac 14 \cos(4x - 1) \:.\: 2 \: dx \\\\
& -\tfrac 14 (2x + 3) \cos(4x - 1) + \tfrac 12 \int \cos(4x - 1) \: dx\\\\
& -\tfrac 14 (2x + 3) \cos(4x - 1) + \tfrac 18 \sin(4x - 1) + C
\end{split}
\end{equation*}