Chapter 5 - Practice Exercises - Page 343: 72

$$\int \frac{(\tan^{-1}x)^2dx}{1+x^2}=\frac{(\tan^{-1}x)^3}{3}+C$$

$$A=\int \frac{(\tan^{-1}x)^2dx}{1+x^2}$$ We set $a=\tan^{-1}x$, which means $$da=\frac{dx}{1+x^2}$$ Therefore, $$A=\int a^2da$$ $$A=\frac{a^3}{3}+C$$ $$A=\frac{(\tan^{-1}x)^3}{3}+C$$