What is the integral of (y-8)/(y^2-y-6) ?

The integral of (y-8)/(y^2-y-6) is 2ln|y+2|-ln|y-3|+C

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integral of (y-8)/(y^2-y-6)

\int \frac{y - 8}{y ^{2} - y - 6} d y

2 ln ∣ y + 2 ∣ - ln ∣ y - 3 ∣ + C

Solution steps

\int \frac{y - 8}{y ^{2} - y - 6} d y

Take the partial fraction of

\frac{y - 8}{y ^{2} - y - 6} : \frac{2}{y + 2} - \frac{1}{y - 3}

= \int \frac{2}{y + 2} - \frac{1}{y - 3} 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 \frac{2}{y + 2} d y - \int \frac{1}{y - 3} d y

\int \frac{2}{y + 2} d y = 2 ln ∣ y + 2 ∣

\int \frac{1}{y - 3} d y = ln ∣ y - 3 ∣

= 2 ln ∣ y + 2 ∣ - ln ∣ y - 3 ∣

Add a constant to the solution

= 2 ln ∣ y + 2 ∣ - ln ∣ y - 3 ∣ + C

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What is the integral of (y-8)/(y^2-y-6) ?
The integral of (y-8)/(y^2-y-6) is 2ln|y+2|-ln|y-3|+C