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