sjg13e
  • sjg13e
Integration by Parts problem
Mathematics
schrodinger
  • schrodinger
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sjg13e
  • sjg13e
\[\int\limits_{}^{}\tan^{-1}xdx\]
anonymous
  • anonymous
try u = arctan x dv = dx
sjg13e
  • sjg13e
\[\let u = \tan^{-1} and du = \frac{ -1 }{ x^2 +1 }dx\]

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sjg13e
  • sjg13e
then dv = dx and v = x
sjg13e
  • sjg13e
so, \[\tan^{-1} \frac{ 1 }{ x } x - \int\limits_{}^{}xdx = \tan^{-1} \frac{ 1 }{ x } x - .5x^2 \]
anonymous
  • anonymous
* du should be positve du 1/(x²+1)
sjg13e
  • sjg13e
why is du positive?
anonymous
  • anonymous
that's the derivative of arctan x. \[\frac{ d }{ dx }\tan^{-1} x=\frac{ 1 }{ 1+x^2 }\]
sjg13e
  • sjg13e
yes but chain rule
sjg13e
  • sjg13e
we need to take the derivative of the inside function, which is 1/x
anonymous
  • anonymous
no. that's an inverse function, not an exponent
sjg13e
  • sjg13e
\[(1/x)\prime = -1/x^2\]
sjg13e
  • sjg13e
i think it's treated as an exponent
sjg13e
  • sjg13e
1/x = x^-1 = -1/2 * x^2
sjg13e
  • sjg13e
ohh I'm sorry!! i forgot to properly write the initial problem. It's supposed to say \[\int\limits_{}^{}\tan^{-1} (\frac{ 1 }{ x }) dx\]
sjg13e
  • sjg13e
i apologize. that's totally my fault
anonymous
  • anonymous
oh ok
sjg13e
  • sjg13e
okay, ahah, so sorry about that! so besides that little mishap, does everything else look right?
anonymous
  • anonymous
yeah it looks good
sjg13e
  • sjg13e
okay thank you!
anonymous
  • anonymous
yw
sjg13e
  • sjg13e
Just for future reference. My previous answer was wrong, here's the correct, complete solution: \[\int\limits_{}^{}\tan^{-1} (\frac{ 1 }{ x })dx = x \tan^{-1}(\frac{ 1 }{ x }) - \int\limits_{}^{}x*\frac{ 1 }{ x^2 +1 }dx \] at this point, you have to do u-subsition in the integral. so, let u = x^2 + 1, which will make .5du = xdx we can bring the - and the .5 out front to give: \[x \tan^{-1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ du }{ u }\] \[ =x \tan^{-1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\ln|u| \] \[= x \tan^{-1} (\frac{ 1 }{ x }) + \ln |x^2 + 1| + C\]

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