#### Answer

$x-\dfrac {1}{x}+c$

#### Work Step by Step

$$\int \dfrac {x^{2}+1}{x^{2}}dx=\int \left( 1+\dfrac {1}{x^{2}}\right) dx=\int \left( 1+x^{-2}\right) dx=x-x^{-1}+c=x-\dfrac {1}{x}+c$$

Published by
Brooks Cole

ISBN 10:
1-43904-957-2

ISBN 13:
978-1-43904-957-0

$x-\dfrac {1}{x}+c$

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