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

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

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integral of 1/(1-z^2)

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

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

Solution steps

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

Use the common integral:

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

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

Add a constant to the solution

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

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