Show that

Let $I=\int x \sec ^{2} x d x$

Taking $x$ as first function and $\sec ^{2} x$ as second function and integrating by parts, we obtain

$\begin{aligned} I &=x \int \sec ^{2} x d x-\int\left\{\left\{\frac{d}{d x} x\right\} \int \sec ^{2} x d x\right\} d x \\ &=x \tan x-\int 1 \cdot \tan x d x \\ &=x \tan x+\log |\cos x|+\mathrm{C} \end{aligned}$