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Homework Help: Indefinite Integration by exchange of variables - Problem 2

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data


    [tex] \int \frac {-1}{x^{2}(1+\frac{1}{x})^{2}} dx [/tex]

    2. Relevant equations

    Same as before:

    [tex] \int f(u)du = F(g(x)) + C [/tex]

    3. The attempt at a solution


    [tex] u = 1 + \frac {1}{x} [/tex]


    [tex] du = -\frac{1}{x^{2}} [/tex]


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

    How does this work out? If u = (1 + 1/x) where did the x^2 go from dx to du?

    Then it goes like this:

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

    According to the book this problem can also be done without substituting anything like so:

    [tex] \int \frac {-1}{x^{2}(1+\frac{1}{x})^{2}} dx = \int \frac {1}{(1+\frac{1}{x})^{2}} * d(1 + \frac{1}{x}) = \frac {(1 + \frac {1}{x})^{-1}}{-1} + C [/tex]

    So my problem here is that I do not understand exactly what is happening, is the reason that we choose u = 1 + 1/x because its derivative is -1/x^2? and that's where the x^2 went in the very first step? Am I on the right track?
  2. jcsd
  3. Apr 26, 2010 #2
    Yes the derivative worked out to allow the x^2 to go away. Maybe you will see it better if you do it like the last one so:

    [tex] du = -\frac{1}{x^{2}}dx [/tex]


    [tex]dx = -x^2du [/tex]

    So if you plug dx into the integral the negatives and the x^2's cancel. That is why choosing your u to be that is so valuable and that is what you want to generally try to do in u substitution.
  4. Apr 26, 2010 #3
    Adding on to the other poster...

    I prefer not to suppress the denominator.

    So, u = 1 + 1/x

    du/dx = (-1/x^2) dx/dx, since dx/dx = 1 you can multiply the rhs by it.

    So, (-x^2) du/dx = dx/dx.

    The dx in your integral can be seen as dx/dx. Substitute it with this value and we are done.

    Basically the same thing as the previous poster has said.
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