
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!

Integration problem

  1. Aug 17, 2004 #1
    After learning a few more techniques at integration and practicing the books thrown me in the deep end, giving me 50 integration questions with usually no clue of what method to use >_<.

    Here's one i'm stuck on, and my working thus far:

    Integrate with respect to x:

    x^2/(x-2) dx
    I rewrote it as: x^2.(x-2)^-1

    I then tried to use integration by parts, thinking that eventually i would be able to bring the x^2 down to 2, and produce an integral i could solve.

    V = x^2, dv/dx = 2x, du = (x-2)^-1, u = ln[x-2].

    x^2ln[x-2] - integral of (2xln[x-2])

    So i thought i would need to use integration by parts again, to solve the above integral and bring the 2x to a 2.

    Integral of (2xln[x-2])
    v = 2x, dv/dx = 2, du = ln[x-2], u = ?, where's where i'm having problems, i've tried using integration by parts again to calculate ln[x-2] but simply can't do it, as the new expression is becoming constantly more difficult.

    I thought about substitution, but i would need to take u = x^2, with du/dx as 2x, however, this would only produce u/(du/dx - 2) dx, which isn't going anywhere.

    Some help would be much appreciated. Thanks.
  2. jcsd
  3. Aug 17, 2004 #2
    Finding the integral of ln(x) is easy. ln(x) is 1 * ln(x). In the integration by parts formula, take the part to be integrated to be "1" and the factor to be differentiated to be "ln(x)" (I think this translates to V = ln(x) and du = 1 in your notation, but I can't be bothered to check it).

    Also, substitution will work for the original integral. Take u = x - 2, then du = dx and x^2 = (u + 2)^2. The integral turns into (u + 2)^2 / u du, which surely is trivial to integrate. Performing polynomial division on x^2/(x - 2) will also work.
    Last edited: Aug 17, 2004
  4. Aug 17, 2004 #3
    Ahh, thanks. So subsitution did work. I totally forgot that i could set u as (x-2) instead of x^2., it makes it even easier as du/dx = 1.
    Silly mistakes, it will come with practice i guess.
  5. Aug 17, 2004 #4
    Okay, i've just done the substitution, using u = x-2 it becomes:

    Integrate: u + 4 + 4.u^-1
    = 0.5u^2 + 4u + 4ln
    = 0.5(x-2)(x-2) + 4(x-2) + 4ln[x-2]
    = 0.5(x^2 - 4x + 4) + 4x - 8 + 4ln[x-2]
    = 0.5x^2 - 2x + 2 + 4x - 8 + 4ln[x-2]
    = 0.5x^2 + 2x - 6 + 4ln[x-2]

    But the book gives the answer as:
    0.5x^2 + 2x + 4ln[x-2].

    Is the book incorrect or have i made a stupid mistake somewhere?
    Thanks for the help.
  6. Aug 17, 2004 #5
    What happens to -6 when you differentiate 0.5x^2 + 2x - 6 + 4ln[x-2]? It becomes zero and vanishes.

    To be absolutely correct, you have to say that the integral of x^2/(x - 2) is 0.5x^2 + 2x + 4ln[x-2] + C, where C is any real number...
  7. Aug 17, 2004 #6
    So the -6 will just effect the constant? As this is integration - not yet solvable differential equations i should just skip the terms without x in as they just effect the constant?
  8. Aug 17, 2004 #7
    Pretty much yes.

    Yes, all constant terms can be removed and replaced with a single constant instead.
  9. Aug 17, 2004 #8
    Thanks very much for the help.

Similar Discussions: Integration problem
  1. Integration problem (Replies: 8)

  2. Integral problem. (Replies: 2)

  3. Integration problem (Replies: 3)

  4. Integration problem (Replies: 4)

  5. Integration problem (Replies: 3)