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Homework Statement
Find:
[tex] \int \frac {-1}{x^{2}(1+\frac{1}{x})^{2}} dx [/tex]
Homework Equations
Same as before:
[tex] \int f(u)du = F(g(x)) + C [/tex]
The Attempt at a Solution
Let
[tex] u = 1 + \frac {1}{x} [/tex]
then
[tex] du = -\frac{1}{x^{2}} [/tex]
so
[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?