Find antiderivative $\int (2x^3+x)(\arctan x)^2dx $

Note that$$\frac{x^3}{1+x^2}=\frac{x^3+x}{1+x^2}-\frac x{1+x^2}=x-\frac x{1+x^2}.$$Can you take it from here?

Let $t=x^2$. Your integral is $\int \frac{1}{6}\frac{tdt}{t+1}=\int (1-\frac{1}{t+1})dt=\frac{t}{6}-\int\frac{1}{t+1}dt$. Can you take it from here?

$$\int\frac{x^3}{1+x^2}dx = \int\left(\frac{x^2}{2(1+x^2)}\right)2xdx$$ Using $u = x^2 \implies du = 2xdx$ $$ = \frac{1}{2}\int\left(1 - \frac{1}{1+u}\right)du = \frac{1}{2}(u - \ln(1+u)) = \boxed{\frac{1}{2}(x^2 - \ln(1+x^2))}$$