## University Calculus: Early Transcendentals (3rd Edition)

$$\int\tan^{-1}xdx=x\tan^{-1}x+\ln|\sec(\tan^{-1}x)|+C$$
$$A=\int\tan^{-1}xdx$$ 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 $$A=x\tan^{-1}x-\int\tan ydy$$ $$A=x\tan^{-1}x-\ln|\sec y|+C$$ $$A=x\tan^{-1}x+\ln|\sec(\tan^{-1}x)|+C$$