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Integrating xln(x^2)?

Can anybody help me understand how to go about integrating this please?
Reply 1
Original post by awilson008
Can anybody help me understand how to go about integrating this please?


Use the Integration by Parts* - take the ln(x^2) as u, as you can't integrate the natural log, and x as v.... It all follows on from there if you know how to do the rest.

Edit: The Integration by Parts* formula is on the C4 Formula Sheet.
Edit 2: *Integration by Parts not Chain Rule!
(edited 10 years ago)
Original post by Ornlu
Use the Chain Rule - take the ln(x^2) as u, as you can't integrate the natural log, and x as v.... It all follows on from there if you know how to do the rest.

Edit: The chain rule formula is on the C4 Formula Sheet.


You mean integration by parts.
Reply 3
Original post by IrrationalNumber
You mean integration by parts.


That's the one; I get confused sometimes :biggrin: You get the gist of what I mean though :redface:
Reply 4
tabular integration by parts - can be written as just 2x(ln(x))
Original post by Hasufel
tabular integration by parts - can be written as just 2x(ln(x))

What is 'tabular integration by parts'?

And yes, it's probably best to recognize log(x2)=2log(x) \log(x^2)=2\log(x) but you can do that before differentiating the logarithm.
Reply 6
make a table of two columns, in the left, the function you want to derive (ln(x)), and in the right, the function to integrate.

then put in the left column the ln(x) with a + sign in front of it, derive it ONCE, and put a -ve sign infront of that. for the left column integrate the 2x, giving x^2, now just multiply the +ln(x) in the left column with the first integral (sorry - edit) in the right one, and take the integral of the first derivative (-1/x) (remember the -ve sign) with the first integral (some x`s will cancel)

check this:

http://www.math.harvard.edu/~knill/teaching/math1a_2012/handouts/39-parts.pdf
(edited 10 years ago)
Okay thanks guys, after a solid hour I've got my head round it :smile:

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