Indefinite Integration by exchange of variables

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SUMMARY

The forum discussion centers on the indefinite integration of the function \(\int \frac {2+z^{-1}}{z^{2}} dz\). The user attempted to solve the integral by substituting \(u = 2 + z^{-1}\) and derived the expression \(-\frac{1}{2}(2+z^{-1})^{2} + C\). However, the book's answer is \(-2z^{-1}-\frac{1}{2}z^{-2} + C\), which is equivalent but presented in a different form. The resolution emphasizes that both answers are correct, highlighting the importance of recognizing equivalent expressions in integration.

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Homework Statement



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

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

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