Determine the integral

  • Thread starter Ocis
  • Start date
  • #1
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Main Question or Discussion Point

Determine the integral (x^2) / (2x^3 - 3)^2

This is probably easier than I expect but I have been attempting it using substitution...?
u = (2x^3 -3)
du/dx = 6x^2
dx = du / 6x^2

x^2 / (u^2) . du / 6x^2 - then I lose it because of the canceling out, and I'm unsure what happens to the x's...

I know the answer is -1/6 (2x^3 - 3) + C, but would appreciate the working I miss out. Or if I am attempting it using the wrong methods.

Many thanks,

Ocis
 

Answers and Replies

  • #2
195
1
you get [tex]\int[/tex]1/6u^2 du, so now integrate 1/u^2 with respect to u
 
  • #3
24
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Thank you for the reply.

Ok so can I say that 1/u^2 = u^-2 ?
if so it would become 1/6 . (u^-1 / -1) = -1/6 (u) ??? = -1/6 (2x^3 - 3) + C

Ok if that is right I think I understand a little more but just get confused with what to do with the x^2 as the numerator...?
 
  • #4
195
1
dx=du/6[tex]x^{2}[/tex]
when you substitute this in the expression the x^2 in the numerator cancels out, so there is in fact no x left in the expression.
 
  • #5
24
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Ok I understand now, thank you very much.
 

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