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Help With Integration by Parts

  1. Apr 26, 2009 #1
  2. jcsd
  3. Apr 27, 2009 #2
    ...I'm sorry but I don't think you understood my questions.
  4. Apr 27, 2009 #3
    1) I don't see what your question is. x^7 = x^4*x^3 because 3 + 4 = 7. This is a basic property of exponents.

    2) Do the substitution w = x^2. Then dw = 2x dx, therefore x dx = 1/2 dw. Using this,

    x^3 * (5 + 3*x^4)^1/2 dx = x^2 * x * (5 + 3*x^4)^1/2 dx = w * x * (5 + 3*w^2)^1/2 dx = w * (5 + 3*w^2)^1/2 x dx = 1/2 w * (5 + 3*w^2)^1/2 du. Then use integral (31) found here: http://www.integral-table.com/" [Broken].
    Last edited by a moderator: May 4, 2017
  5. Apr 27, 2009 #4
    For the expression they split out:
    dv = x3(5+3x4)1/2

    You need to find an integral of this. An integral of this would look something along the lines of
    (5 + 3x4)3/2
    Differentiating (5 + 3x4)3/2 would give you
    12x3 * 3/2 (5 + 3x4)1/2
    18x3(5 + 3x4)1/2
    so to bring it in line with the expression you have, the integral needs a constant in the form of 1/18 meaning
    integrates to
    If you don't follow the logic of this, revisit some more basic integration.

    This also explains the logic behind splitting x7 into x4 and x3. It gives you the correct power of x to multiply the bracket with in order to make an integration of the bracket easier (without doing this, you would have:
    u = x7
    dv = (5 + 3x4)1/2
    v = 1/18 * x-3(5 + 3x4)3/2
    which just isn't as nice to work with (and the factor of x-3 should end up cancelling out in the next step anyway)
    Last edited: Apr 27, 2009
  6. Apr 27, 2009 #5
    I appreciate the help. I'm trying to learn Calc II by myself, and some of these forms of integration elude me. I need to see one example, and understand one example of each type before I "get it," and this is the first time I've had to integrate something like this. Just yesterday I began grasping the integration of the "brothers" (for lack of a better word) of e and lnx...now I'm trying to understand roots.

    A couple of questions:

    Do you just drop the x3? Where does the /(3/2) go to?

    Basically, why isn't the integral of x3(5+3x)1/2.....x4 / 4 (2/3)(5+3x)3/2?

    And when you differentiate again...how does the 1/12 suddenly get reintroduced?

    So...once you see that x3 = (1/12), you use that to simplify all x3?



    I think I just saw something. Do you take the derivative of what's under the square root and multiply it by whats outside...resulting in x3 / 12x3...equalling 1/12? If this is the case, how does the 1/12 disappear and then reappear?
    Last edited: Apr 27, 2009
  7. Apr 27, 2009 #6
    Ok, remember that integration is the reverse of differentiation. If you want to differentiate a formula such as:
    v = (5 + 3x4)3/2
    Then the differentiation really needs to be done by substitution as such:

    U = 5 + 3x4
    v = U3/2

    dv/dx = (dU/dx)(dv/dU)

    Now, differentiating,
    dv/dU = 3/2U1/2 (a pretty standard differentiation)
    dU/dx = 12x3


    dv/dx = 12x3 * 3/2U1/2
    and substituting U back in and simplifying
    dv/dx = 18x3(5 + 3x4)1/2

    so, integrating is the reverse of this. Therefore if you have a formula such as
    dv = x3(5 + 3x4)1/2
    you can see it would integrate to something like
    v = (5 + 3x4)3/2
    but in this case, we don't have the factor of 18 at the front. So to get the correct integral (ignoring the constant of integration here) you need to divide your final answer by this factor to correct it, arriving at an integral of:
    1/18(5 + 3x4)3/2

    Now, with your integration by parts, you can either split it along the obvious bounds of
    dv = (5 + 3x4)1/2
    and procede simply like this, or you can recognise that you can 'borrow' x3 from x7 to give you
    dv=x3(5 + 3x4)1/2
    and this gives you a simpler v to work with when you integrate it. Either way should produce the same answer in this case, however it is useful to learn to recognise these forms as sometimes a split like that is the only way to correctly integrate, or provides a much easier route to the final answer.

    I hope that makes things clearer.

    Also, if you are self-teaching and want to learn (one method) of how integration works, I'd suggest going for something like the book here:
    This is an approach to calculus from first principles using the fairly easily understood 'infinitesimal' approach with hyperreal numbers (basically real numbers with the concept of infinitely small and infinitely big numbers introduced). With a first principles approach like this, all these integration and differentiation forms fall out algebraically which makes them much easier to understand and remember in the future :)
    Last edited: Apr 27, 2009
  8. Apr 27, 2009 #7
    That was an amazing help. Thanks so much.
  9. Apr 27, 2009 #8
    Oh, and for a better word over 'brothers' for ex and ln(x), there is always the word inverse ;) ln(x) is the inverse of ex
  10. Apr 27, 2009 #9
    Hahaha no I meant like...e^x^2 and the like for brothers. I must have asked a really stupid question because I'm being really underestimated! I did have the highest grade in my Calc I course and was the only person to be exempted from the exam. I'm not a complete dolt! I just am teaching myself Calc II and III before next semester because I'm going to have a difficult math sequence (doubling up to catch up on my major) and it can be hard to understand everything without a teacher's explanation.
  11. Apr 27, 2009 #10
    Just saw the book. Friggen amazing. I'm going to get to reading ASAP.

    I think part of my problem is just not understanding the background info of the integral. I can do u-substitution, partial integration, and a decent amount of integration by parts...but when you mix the integration by parts with needed u-substitutions I can get lost. But then again, I haven't seen this stuff before...but hopefully this book will familiarize me with things so I know WHY to do _________ instead of just how.
  12. Apr 27, 2009 #11
    Heh :) no problem. The reason I'm not (nor ever would be)a teacher is because I tend to go either too complicated or too simple with explanations, I guess this was one of them :)

    I'm not familiar with american courses either though... I remember doing this stuff in my Further Maths A-Level about 8 years ago now... and then my degree where I never touched the stuff and now I'm interested in maths and physics again. That book was a fairly recent find of mine and although I haven't read all of it yet I'm happier with my knowledge of calculus just knowing there are nice algebraic methods to work everything out and rough ideas of what they are :)
  13. Apr 27, 2009 #12
    Wow. Okay so in calc I we breezed through everything calculus related and finished with some linear algebra. We technically finished calc I in only 1.5 months. So...fearing that I had forgotten aspects of u-substitution, I looked back over things. Apparently, that was what I was missing in understanding this problem.
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