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

The integral of 6/(y^2-2y-8) is -ln|y+2|+ln|y-4|+C

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

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

- ln ∣ y + 2 ∣ + ln ∣ y - 4 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{y ^{2} - 2 y - 8} d y

Take the partial fraction of

\frac{1}{y ^{2} - 2 y - 8} : - \frac{1}{6 ( y + 2 )} + \frac{1}{6 ( y - 4 )}

= 6 \cdot \int - \frac{1}{6 ( y + 2 )} + \frac{1}{6 ( y - 4 )} 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

= 6 (- \int \frac{1}{6 ( y + 2 )} d y + \int \frac{1}{6 ( y - 4 )} d y)

\int \frac{1}{6 ( y + 2 )} d y = \frac{1}{6} ln ∣ y + 2 ∣

\int \frac{1}{6 ( y - 4 )} d y = \frac{1}{6} ln ∣ y - 4 ∣

= 6 (- \frac{1}{6} ln ∣ y + 2 ∣ + \frac{1}{6} ln ∣ y - 4 ∣)

Simplify

6 (- \frac{1}{6} ln ∣ y + 2 ∣ + \frac{1}{6} ln ∣ y - 4 ∣) : - ln ∣ y + 2 ∣ + ln ∣ y - 4 ∣

= - ln ∣ y + 2 ∣ + ln ∣ y - 4 ∣

Add a constant to the solution

= - ln ∣ y + 2 ∣ + ln ∣ y - 4 ∣ + C

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What is the integral of 6/(y^2-2y-8) ?
The integral of 6/(y^2-2y-8) is -ln|y+2|+ln|y-4|+C