Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 65

$$\int \frac{dy}{(\tan^{-1}y)(1+y^2)}=\ln|\tan^{-1}y|+C$$

$$A=\int \frac{dy}{(\tan^{-1}y)(1+y^2)}$$ We set $u=\tan^{-1}y$ Then $$du=\frac{dy}{1+y^2}$$ Therefore, $$A=\int \frac{1}{u}du=\ln|u|+C$$ $$A=\ln|\tan^{-1}y|+C$$