$\displaystyle \int{x(arctanx)^2} dx $
No work to show so far, not sure which substitutions would be the best to make here.
If I have done this correctly,
$\displaystyle du=2(\arctan x)\frac{1}{1+x^2}dx, ~v=\frac{1}{2}x^2 $
the integral is then $\displaystyle \frac{1}{2}x^2(arctan x)^2 - \int \frac{1}{2}x^22(\arctan x)\frac{1}{1-x^2}dx $
I am then stuck on what to do with that integral..should I make the substitution u = arctan x and then do parts again?
$\displaystyle \frac{1}{2}x^2(\arctan x)^2 - \int x^2(\arctan x)\frac{1}{1+x^2}dx $
$\displaystyle \frac{1}{2}x^2(\arctan x)^2 - \int \left(1 - \frac{1}{1+x^2}\right)\arctan{x} \, dx $
$\displaystyle \frac{1}{2}x^2(\arctan x)^2 - \int \arctan{x} \, dx + \int \frac{\arctan{x}}{1+x^2} \, dx $
continue ...
The red one should be "+" not "-".=Em Yeu Anh;440924]If I have done this correctly,
the integral is then $\displaystyle \frac{1}{2}x^2(arctan x)^2 - \int \frac{1}{2}x^22(\arctan x)\frac{1}{1{\color{red}-}x^2}dx $
Also the 2`s will cancel.
and the integral will be:
$\displaystyle \int tan^{-1}(x) \frac{x^2}{x^2+1} dx$
Clearly: $\displaystyle \frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}$ <--- Do you know why?
Substituting this in the integral:
$\displaystyle \int tan^{-1}(x) (1-\frac{1}{x^2+1}) dx$
$\displaystyle =\int tan^{-1}(x)dx-\int \frac{tan^{-1}(x)}{x^2+1}dx$
This first one can be solved using integration by parts.
The second one can be solved by using a U-Substitution.