<s>[INST] Find antiderivative $\int (2x^3+x)(\arctan x)^2dx $ 
Find antiderivative $$\int (2x^3+x)(\arctan x)^2dx $$

My try:
$$\int (2x^3+x)(\arctan x)^2dx =(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int \frac{2\arctan x}{1+x^2}(\frac{1}{2}x^4+\frac{1}{2}x^2)dx=(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int (\arctan x) (x^2)dx= (\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\arctan x\cdot \frac{1}{3}x^3-\int (\frac{1}{1+x^2}) (\frac{1}{3}x^3)dx$$And then I don't know how I can find $\int (\frac{1}{1+x^2}) (\frac{1}{3}x^3)dx$. Can you help me with it?
 [/INST] 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?
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