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Homework Help: I'm hopeless at integration

  1. Jan 15, 2012 #1
    Hi all
    I'm a 24 year old college student and am learning calculus over my holiday break. I studied a subject last semester which taught calculus up to partial derivatives, but it was VERY poorly taught (to the point where the university is evaluating the results of all students of the subject). I absolutely love math and I would like to study calculus and its applications in my spare time (particularly astronomy applications)

    I am struggling with a few things, in particular integration by substitution. The professor showed an example of a very basic function but did not explain when this method is used, or why. From what I can gather, it is used when the function also contains its own derivative.
    I had enlisted the help of a maths tutor because I hadn't studied math in about 7 years (and never any calculus).
    Anyway here is my question.

    1. The problem statement, all variables and given/known data

    Integration by substitution
    f(x) = 2x/(5+x^2)

    2. Relevant equations

    3. The attempt at a solution

    Now, if we allow u(x) = 5+x^2, then du/dx = 2x. We move 2 out of the integral, but what happens to the x as the numerator? I am to understand it equals 1, as the derivative of u is removed from the integral. The integral is now 1/u which is ln (5+x^2) + C.

    Now, I don't understand why exactly the derivative is gone! My maths tutor had shown me a method, but I don't understand exactly how it works. The method is as follows:
    f(x)= 2x/(5+x^2) dx
    u(x)= 5+x^2, du/dx= x --> du = x^2/2
    =2 ∫ 1/(5+x^2) dx^2/2
    =2/2 ∫ 1/(5+x^2) dx^2
    = ∫ 1/(5+x^2) dx^2+5
    = ∫ 1/u du
    = ln|u|
    =ln(5+x^2) + C

    His reasoning was that we had to match the dx on the RHS with the function, and then balance it out with the coefficient outside the integral sign. I've never heard of this, and my professor or tutors at university never mentioned any such method. Actually, they haven't even explained the meaning of 'dx' on the RHS so I am really lost.
    I know this is a long post but I appreciate your time and would really appreciate any help.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 15, 2012 #2


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    Homework Helper

    This method of substitution is especially easy when the integrand is the product of a composite function f(u(x)) and the derivative of u(x) =u'.
    [tex]\int f(u(x))u'(x) dx[/tex]

    You can writhe the derivative as [itex]u'=\frac{du}{dx}[/itex] and then [itex]u' dx=\frac{du}{dx} dx=du[/itex], so the integral becomes

    [tex]\int f(u(x)) \frac{du}{dx} dx=\int f(u)du[/tex]

    The example integral is

    [tex]\int \frac{2x}{5+x^2}dx=\int \frac{1}{5+x^2}2x dx[/tex]

    Let be u=5+x^2. u'=2x, so you can substitute 2x by u', u'dx by du.

    [tex]\int \frac{1}{5+x^2}(2x dx)=\int \frac{1}{u}(u'dx)=\int \frac{1}{u}du[/tex]

    Integrate with respect to u, you get ln|u| +C, and substitute 5+x^2 for u: the integral is ln|5+x^2|+C.

  4. Jan 15, 2012 #3
    Here is an alternative to what ehild wrote. It's basically the same thing though.

    The x in the numerator was cancelled when you substituted in du for dx. Since we can't integrate something with u when there is still a dx in there, we replace it with something that contains du. You had it right, du/dx = 2x. But that also means that dx = du/2x, right?

    We have this:
    [tex] \int \frac{2x}{5+x^2}dx [/tex]
    Then substituting in u for 5+x^2 and du/2x for dx we get this:
    [tex] \int \frac{2x}{u} \frac{du}{2x}= \int \frac{1}{u}du [/tex]
    See what happened? When I replaced the dx with du/2x, stuff cancels. This is how I learned it, but I think ehild's response is a little more proper.
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