Answer
$$\int\frac{(\arctan x)^2}{x^2+1}dx=\frac{(\arctan x)^3}{3}+C$$
Work Step by Step
$$A=\int\frac{(\arctan x)^2}{x^2+1}dx$$
Let $u=\arctan x=\tan^{-1}x$.
We would have $du=\frac{1}{x^2+1}dx$.
Substitute into $A$, we have $$A=\int u^2du$$ $$A=\frac{u^3}{3}+C$$ $$A=\frac{(\arctan x)^3}{3}+C$$
