# Indefinite Integration by exchange of variables

1. Apr 28, 2010

### Asphyxiated

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

$$\int \frac {2+z^{-1}}{z^{2}} dz$$

3. The attempt at a solution

Let:

$$u = 2 +z^{-1}$$

$$du = -z^{-2} dz$$

$$dz = -z^{2} du$$

so now its

$$\int \frac {u}{z^{2}} (-z^{2}) du$$

$$\int \frac {(u)(-z^{2})}{z^{2}} du$$

$$\int (u)(-1) du$$

and then the antiderivative of u*(-1) is

$$-\frac{1}{2}(2+z^{-1})^{2} + C$$

right? The answer in the book is:

$$-2z^{-1}-\frac{1}{2}z^{-2} + C$$

I don't see anywhere that I went wrong....

2. Apr 28, 2010

### happyg1

expand your parenthesis and put the -2 that results in with the constant. You have the same answer, just in a different form.