anonymous
  • anonymous
integral of x*arctan(x)dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
1/2 (-x+(1+x^2) tan^(-1)(x))+C
anonymous
  • anonymous
i need to know how to get there not just the answer I'm studying for a test not just trying to get homework done.
anonymous
  • anonymous
parts for this one but it is a pain

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More answers

anonymous
  • anonymous
\[u=\tan^{-1}(x), du =\frac{1}{x^2+1}, dv=x, v=\frac{x^2}{2}\] gives \[\frac{x^2}{2}\tan^{-1}(x)-\int \frac{x^2}{2(x^2+1)}dx\] but second integral is not so fun
anonymous
  • anonymous
right there is where I got stuck
saifoo.khan
  • saifoo.khan
@satellite73 please kick the spammer from the chat. Thanks
anonymous
  • anonymous
oh actually the second one is not so bad as i though, because \[\frac{x^2}{x^2+1}=1-\frac{1}{x^2+1}\] and the integral of the second piece is back to arctan
anonymous
  • anonymous
I don't see how \[x^{2}/x^{2}+1=1-1/x^{2}+1\]
anonymous
  • anonymous
divide
anonymous
  • anonymous
whenever the degree of the numerator is the same as the denominator (or larger) it is usually the right thing to do
anonymous
  • anonymous
i see it now thanks
anonymous
  • anonymous
or you can use a trick, namely \[\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}\] but that trick really works best if you already know the answer
anonymous
  • anonymous
That is actually how I saw it in my head when it clicked

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