What is the integral of 9/(z^2-12z) ?

The integral of 9/(z^2-12z) is 9(-1/12 ln|z|+1/12 ln|z-12|)+C

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integral of 9/(z^2-12z)

\int \frac{9}{z ^{2} - 12 z} d z

9 (- \frac{1}{12} ln ∣ z ∣ + \frac{1}{12} ln ∣ z - 12 ∣) + C

Solution steps

\int \frac{9}{z ^{2} - 12 z} d z

Take the constant out:

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

= 9 \cdot \int \frac{1}{z ^{2} - 12 z} d z

Take the partial fraction of

\frac{1}{z ^{2} - 12 z} : - \frac{1}{12 z} + \frac{1}{12 ( z - 12 )}

= 9 \cdot \int - \frac{1}{12 z} + \frac{1}{12 ( z - 12 )} d z

Apply the Sum Rule:

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

= 9 (- \int \frac{1}{12 z} d z + \int \frac{1}{12 ( z - 12 )} d z)

\int \frac{1}{12 z} d z = \frac{1}{12} ln ∣ z ∣

\int \frac{1}{12 ( z - 12 )} d z = \frac{1}{12} ln ∣ z - 12 ∣

= 9 (- \frac{1}{12} ln ∣ z ∣ + \frac{1}{12} ln ∣ z - 12 ∣)

Add a constant to the solution

= 9 (- \frac{1}{12} ln ∣ z ∣ + \frac{1}{12} ln ∣ z - 12 ∣) + C

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What is the integral of 9/(z^2-12z) ?
The integral of 9/(z^2-12z) is 9(-1/12 ln|z|+1/12 ln|z-12|)+C