Integration by parts?

  • Thread starter mpgcbball
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  • #1
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im having a bit of trouble, can anyone help me integrate arctan(1/x) using integration by parts?
thanks
 

Answers and Replies

  • #2
dextercioby
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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]
 
  • #3
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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
 
  • #4
132
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Take it to be 1* arctan(...) then your dv will be 1.
 
  • #5
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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(???)
 
  • #6
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ooooh i get it!! thank you
 
  • #7
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im still getting stuck at xarctan(1/x)-int(-x/x^2+1)
 
  • #8
dextercioby
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Can you then integrate

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

?
 
  • #9
132
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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]
 
  • #10
184
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what is the derivative of [tex]\ln(x^2 + 1) [/tex] ?
 

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