Integration by parts?

mpgcbball · Feb 15, 2007

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

dextercioby · Feb 15, 2007

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]

mpgcbball · Feb 15, 2007

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

trajan22 · Feb 15, 2007

Take it to be 1* arctan(...) then your dv will be 1.

mpgcbball · Feb 15, 2007

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(???)

mpgcbball · Feb 15, 2007

ooooh i get it!! thank you

mpgcbball · Feb 15, 2007

im still getting stuck at xarctan(1/x)-int(-x/x^2+1)

dextercioby · Feb 15, 2007

Can you then integrate

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

?

trajan22 · Feb 15, 2007

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]

theperthvan · Feb 15, 2007

what is the derivative of [tex]\ln(x^2 + 1) [/tex] ?

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Integration by parts?

mpgcbball

dextercioby

mpgcbball

trajan22

mpgcbball

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mpgcbball

dextercioby

trajan22

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