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Popular >

integral of 1/(z-1)

Solution

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

Solution

ln ∣ z - 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = z - 1

= ln ∣ z - 1 ∣

Add a constant to the solution

= ln ∣ z - 1 ∣ + C

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Frequently Asked Questions (FAQ)

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