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

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

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integral of 1/(z^2-z)

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

- ln ∣ z ∣ + ln ∣ z - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{z ^{2} - z} : - \frac{1}{z} + \frac{1}{z - 1}

= \int - \frac{1}{z} + \frac{1}{z - 1} 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

= - \int \frac{1}{z} d z + \int \frac{1}{z - 1} d z

\int \frac{1}{z} d z = ln ∣ z ∣

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

= - ln ∣ z ∣ + ln ∣ z - 1 ∣

Add a constant to the solution

= - ln ∣ z ∣ + ln ∣ z - 1 ∣ + C

\int_{1}^{7} \frac{1}{9 + ( x + 2 ) ^{2}} d x

\int (2 x^{3} - 3 x + 5) d x

\int x cot (x^{2}) d x

y^{^{'}} + y = e^{- t}

n = 1 \sum \infty \frac{n - 5}{n} + 5

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