Answer
$$\int\sec^{-1}xdx=x\sec^{-1}x-\ln\Big|\sec(\sec^{-1}x)+\tan(\sec^{-1}x)\Big|+C$$
Work Step by Step
$$A=\int\sec^{-1}xdx$$
Take $y=f^{-1}(x)=\sec^{-1}x$
So $x=f(y)=\sec y$
Applying the formula $\int f^{-1}(x)dx=xf^{-1}x-\int f(y)dy$, we have
$$A=x\sec^{-1}x-\int\sec ydy$$ $$A=x\sec^{-1}x-\ln|\sec y+\tan y|+C$$ $$A=x\sec^{-1}x-\ln\Big|\sec(\sec^{-1}x)+\tan(\sec^{-1}x)\Big|+C$$
