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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?

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    $\begingroup$ Do the division ($x^3/(1+x^2)$). $\endgroup$ – David Mitra May 13 '19 at 18:02
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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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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?

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

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