# 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?

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

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