$\displaystyle \int{x(arctanx)^2} dx $

No work to show so far, not sure which substitutions would be the best to make here.

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- Jan 17th 2010, 09:11 PMEm Yeu AnhIntegrating by Parts
$\displaystyle \int{x(arctanx)^2} dx $

No work to show so far, not sure which substitutions would be the best to make here. - Jan 17th 2010, 09:40 PMJhevon
- Jan 18th 2010, 03:44 PMEm Yeu Anh
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? - Jan 18th 2010, 04:02 PMskeeter
$\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 ... - Jan 18th 2010, 04:05 PMGeneralQuote:

=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 $

**The red one should be "+" not "-".**

**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.**