What is the integral of (y^2-1)/y ?

The integral of (y^2-1)/y is (y^2)/2-ln|y|+C

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integral of (y^2-1)/y

\int \frac{y ^{2} - 1}{y} d y

\frac{y ^{2}}{2} - ln ∣ y ∣ + C

Solution steps

\int \frac{y ^{2} - 1}{y} d y

Expand

\frac{y ^{2} - 1}{y} : y - \frac{1}{y}

= \int y - \frac{1}{y} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y d y - \int \frac{1}{y} d y

\int y d y = \frac{y ^{2}}{2}

\int \frac{1}{y} d y = ln ∣ y ∣

= \frac{y ^{2}}{2} - ln ∣ y ∣

Add a constant to the solution

= \frac{y ^{2}}{2} - ln ∣ y ∣ + C

What is the integral of (y^2-1)/y ?
The integral of (y^2-1)/y is (y^2)/2-ln|y|+C