Integrate: $\int x(\arctan x)^{2}dx$

Question

I'm not sure how to start I think we have to use integration by parts $$\int x(\arctan x)^{2}dx$$

Answer 1

With $\tan(u)=x$, $$ \begin{align} \int x\arctan(x)^2\mathrm{d}x &=\int\tan(u)\,u^2sec^2(u)\,\mathrm{d}u\\ &=\frac12\int u^2\,\mathrm{d}\sec^2(u)\\ &=\frac12u^2\sec^2(u)-\int u\sec^2(u)\,\mathrm{d}u\\ &=\frac12u^2\sec^2(u)-\int u\,\mathrm{d}\tan(u)\\ &=\frac12u^2\sec^2(u)-u\tan(u)+\int\tan(u)\,\mathrm{d}u\\ &=\frac12u^2\sec^2(u)-u\tan(u)-\log|\cos(u)|+C\\ &=\frac12(1+x^2)\arctan(x)^2-x\arctan(x)+\frac12\log(1+x^2)+C \end{align} $$

Answer 2

Using integration by parts, we get

$$\begin{align} \int dx \: x (\arctan{x} )^2 &= \frac{1}{2} x^2 (\arctan{x} )^2 - \int dx \: \frac{x^2}{1+x^2} \arctan{x} \\ &= \frac{1}{2} x^2 (\arctan{x} )^2 - \int dx \: \arctan{x} + \int dx \: \frac{\arctan{x}}{1+x^2} \\ &= \frac{1}{2} x^2 (\arctan{x} )^2 - x \arctan{x} + \int dx \: \frac{x}{1+x^2} + \frac{1}{2} (\arctan{x})^2 \\ &= \frac{1}{2} [(1+x^2) (\arctan{x})^2 + \log{(1+x^2)}] - x \arctan{x} + C \end{align}$$

Answer 3

Yes use integration by parts $$\int x(\arctan x)^{2}dx = (\arctan x)^2 \int x dx - \int \frac{2 \arctan x}{1 + x^2} \left( \int x dx \right) dx \\ = \frac{(\arctan x)^2 x^2}{2} - \int \frac{2 \arctan x}{1 + x^2} \frac{x^2}{2} dx $$

For the latter part see here.