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

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

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integral of 1/(1-x^2)

integral

\frac{1}{1 - x ^{2}}

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

Solution steps

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

Use the common integral:

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

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

Add a constant to the solution

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

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