Answer
$$\int\frac{(\arctan x)^2}{x^2+1}dx=\frac{(\arctan x)^3}{3}+c$$
Work Step by Step
To evaluate the integral $$\int \frac{(\arctan x)^2}{x^2+1}dx$$
we will use substitution $\arctan x=t$ which gives us $\frac{dx}{x^2+1}=dt$. Putting this into the integral we get:
$$\int\frac{(\arctan x)^2}{x^2+1}dx=\int t^2dt=\frac{t^3}{3}+c$$
where $c$ is arbitrary constant. Now we have to express solution in terms of $x$:
$$\int\frac{(\arctan x)^2}{x^2+1}dx=\frac{t^3}{3}+c=\frac{(\arctan x)^3}{3}+c$$
