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

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

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integral of 1/(-w^2+1)

\int \frac{1}{- w ^{2} + 1} d w

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

Solution steps

\int \frac{1}{- w ^{2} + 1} d w

Use the common integral:

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

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

Add a constant to the solution

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

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