Answer
$$ \int \sec^{-1}xdx = x\sec^{-1}x-\ln | x+\sqrt{x^2-1}| +C$$
Work Step by Step
Given
$$ \int \sec^{-1}xdx $$
Use integration by parts , let
\begin{aligned}
u&= \sec^{-1}x\ \ \ \ \ \ \ &dv&= dx\\
du&= \frac{dx}{x\sqrt{x^2-1}}\ \ \ \ \ \ \ & v&= x
\end{aligned}
Then
\begin{aligned}
\int \sec^{-1}xdx &= x\sec^{-1}x-\int \frac{dx}{\sqrt{x^2-1}}
\end{aligned}
To evaluate $\int \frac{dx}{\sqrt{x^2-1}}$, let
$$x=\sec \theta \ \ \ \ \ dx=\sec \theta \tan \theta d\theta $$
Then
\begin{aligned}
\int \frac{dx}{\sqrt{x^2-1}}&=\int \frac{\sec \theta \tan \theta d\theta}{\sqrt{\sec^2\theta -1}}\\
&=\int \sec \theta d\theta \\
&=\ln |\sec \theta +\tan \theta |+c\\
&= \ln | x+\sqrt{x^2-1}|
\end{aligned}
Hence
\begin{aligned}
\int \sec^{-1}xdx &= x\sec^{-1}x-\ln | x+\sqrt{x^2-1}| +C
\end{aligned}
