<s>[INST] Calculus 2: Integration by Parts Stuck on Integral of Product With ArcTan Inside I'm stuck on the following problem: 
$$\int_0^{1/3} y \tan^{-1}(3y)\,dy$$
I think my last line in my work below is correct but I don't know what to do beyond that.  
A step through of the problem would be appreciated.  

 [/INST] Let $x = 3y$ then  $ 3dy = dx $,
$$\displaystyle \int_0 ^{\frac {1}{3}} y \ \tan^{-1}(3y) dy = \frac {1}{9} \int_0 ^1 x\ \tan^{-1}(x) dx$$ 
Use ILATE
$\displaystyle \Rightarrow  \frac {1}{18} [x^2 \tan^{-1}(x) |_0 ^1  - \int_0 ^1 \frac {x^2}{1+x^2} dx ]   $
$\displaystyle \Rightarrow  \frac {1}{18} [x^2 \tan^{-1}(x) |_0 ^1  - \int_0 ^1 \frac {x^2}{1+ x^2} dx ]   $
$\displaystyle \Rightarrow  \frac {1}{18} [x^2 \tan^{-1}(x) |_0 ^1  - ( \int_0 ^1 dx - \int_0 ^1  \frac {1}{1+ x^2} dx )] $
$\displaystyle \Rightarrow  \frac {1}{18} [x^2 \tan^{-1}(x) - x + \tan^{-1}(x) ]_0 ^1   $
$\displaystyle \Rightarrow  \frac {1}{18} [(x^2 + 1) \tan^{-1}(x) - x  ]_0 ^1   
 = \frac {\pi}{36} - \frac {1}{18}$
Please tell if there are errors in this, because I haven't studied calculus yet.
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