Integral by Parts
\(\displaystyle \int u \:.\: dv = u \:.\: v - \int v \: du\)
Example
\(\int (2x + 3) \sin(4x - 1)\:dx\)
\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 & = \frac 14 \int \sin (4x - 1) \: d(4x - 1) \\\\ v & = - \frac 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) \:.\: - \frac 14 \cos(4x - 1) - \int - \frac 14 \cos(4x - 1) \:.\: 2 \: dx \\\\ & -\frac 14 (2x + 3) \cos(4x - 1) + \frac 12 \int \cos(4x - 1) \: dx\\\\ & \bbox[5px, border: 2px solid magenta] {-\frac 14 (2x + 3) \cos(4x - 1) + \frac 18 \sin(4x - 1) + c} \end{split} \end{equation*}
Exercise