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Integration by parts?

  1. Feb 15, 2007 #1
    im having a bit of trouble, can anyone help me integrate arctan(1/x) using integration by parts?
  2. jcsd
  3. Feb 15, 2007 #2


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    Here's the first line

    [tex] \int \arctan \frac{1}{x} {} dx =x\arctan \frac{1}{x}-\int x \frac{-\frac{1}{x^2}}{1+\frac{1}{x^2}} {} dx [/tex]
  4. Feb 15, 2007 #3
    im not really understanding how to get that line. i dont know what to assign as u and dv. the xarctan1/x is the part that confuses me because i dont see where the x comes from
  5. Feb 15, 2007 #4
    Take it to be 1* arctan(...) then your dv will be 1.
  6. Feb 15, 2007 #5
    that doesnt make any sense to me, but thankyou for trying to help. i dont know what "it" is referring to that im supposed to be taking as 1*arctan(???)
  7. Feb 15, 2007 #6
    ooooh i get it!! thank you
  8. Feb 15, 2007 #7
    im still getting stuck at xarctan(1/x)-int(-x/x^2+1)
  9. Feb 15, 2007 #8


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    Can you then integrate

    [tex] \int \frac{x}{x^2 +1} {}dx [/tex]

  10. Feb 15, 2007 #9
    I havent actually done it but if you are right up to that point then it appears that all you have to do is make a simple u substitution to solve the integral.
    [tex] \int \frac{x}{x^2 +1} {}dx [/tex]
  11. Feb 15, 2007 #10
    what is the derivative of [tex]\ln(x^2 + 1) [/tex] ?
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