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

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

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integral of 1/(z^2-2z)

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

- \frac{1}{2} ln ∣ z ∣ + \frac{1}{2} ln ∣ z - 2 ∣ + C

Solution steps

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

Complete the square

z^{2} - 2 z : (z - 1)^{2} - 1

= \int \frac{1}{( z - 1 ) ^{2} - 1} d z

Apply u-substitution

: \int \frac{1}{u ^{2} - 1} d u

= \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= - \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = z - 1

= - (\frac{ln ∣ z - 1 + 1 ∣}{2} - \frac{ln ∣ z - 1 - 1 ∣}{2})

Simplify

- (\frac{ln ∣ z - 1 + 1 ∣}{2} - \frac{ln ∣ z - 1 - 1 ∣}{2}) : - \frac{1}{2} ln ∣ z ∣ + \frac{1}{2} ln ∣ z - 2 ∣

= - \frac{1}{2} ln ∣ z ∣ + \frac{1}{2} ln ∣ z - 2 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ z ∣ + \frac{1}{2} ln ∣ z - 2 ∣ + C

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