Answer
\begin{aligned}
\int\sec ^{-1} x d x= x \sec ^{-1} x-\ln |x+\sqrt{x^{2}-1}|+C
\end{aligned}
Work Step by Step
Given $$
\int \sec ^{-1} x d x
$$
Use integration by parts:
$$ u=\sec ^{-1} x \Rightarrow du= \frac{1}{x\sqrt {x^2-1}}dx $$ $$ dv= dx \Rightarrow v= x $$ So, we get
\begin{aligned}
I&=\int\sec ^{-1} x d x\\
&=uv - \int vdu\\
&=x\sec ^{-1} x-\int \frac{x}{x\sqrt {x^2-1}}dx\\
&=x\sec ^{-1} x-\int \frac{1}{\sqrt {x^2-1}}dx\\
&=x\sec ^{-1} x-\ln |x+\sqrt{x^{2}-1}|+C
\end{aligned}
Check:
$$
\begin{aligned} \frac{d}{d x}[\ln |x+\sqrt{x^{2}-1}|] &=\frac{1}{x+\sqrt{x^{2}-1}} \cdot\left(1+\frac{1}{2 \sqrt{x^{2}-1}} \cdot 2 x\right) \\ &=\frac{1}{x+\sqrt{x^{2}-1}} \cdot\left(1+\frac{x}{\sqrt{x^{2}-1}}\right) \\ &=\frac{1}{x+\sqrt{x^{2}-1}} \cdot\left(\frac{\sqrt{x^{2}-1}+x}{\sqrt{x^{2}-1}}\right) \\ &=\frac{1}{\sqrt{x^{2}-1}} \end{aligned}
$$
