(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find:

[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

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?

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

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