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

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

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

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

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

Solution steps

\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}

Add a constant to the solution

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

\frac{( x ^{8} )}{8 ( 1 - x ^{2} ) ^{4}}

\frac{d}{d x} (\frac{1}{( 9 + x ) ^{2}})

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\frac{d}{d x} (ln (6 - 5 x))

t e^{, at}

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