Indefinite integral with a rational function

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SUMMARY

The discussion centers on finding the indefinite integral of the rational function \(((y^2-1)/y)^2 dy\). The user initially attempted to simplify the expression to \((y - 1/y)^2 dy\) and calculated the antiderivative as \(y^3/3 - 2y + 1/y + C\). However, the correct solution, as confirmed by Wolfram's integral calculator, is \(y^3/3 - 2y - 1/y + C\). The user identified a sign error in the antiderivative of the term \(\int \frac{1}{y^2}dy\), which should yield \(-\frac{1}{y} + C\).

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Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of common errors in solving indefinite integrals.

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EDIT: Problem found. This thread can now be ignored.

Homework Statement



Find the indefinite integral.

Homework Equations



((y^2-1)/y)^2 dy

The Attempt at a Solution



I've attempted a few things. I first attempted to split the statement inside the outer parentheses into two fractions;

(y^2/y - 1/y)^2 dy

Then to reduce it to,

(y - 1/y)^2 dy

And then foil it and take an antiderivative. This comes out to

Foiled: y^2 - 2 + 1/y^2 dy

And then the antiderivative:

y^3/3 - 2y + 1/y + C

But I appear to still be incorrect. According to Wolfram's integral calculator, the solution is

y^3/3 - 2y - 1/y

I'm close. I'm apparently missing a sign somewhere, and I can't seem to find where it is, and I don't feel comfortable plugging in the answer until I know how I got there.
 
Last edited:
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Your error is just when you are finding the anti-derivative of the last term:

\int \frac{1}{y^2}dy=-\frac{1}{y}+C
 

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