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

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

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

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

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

Solution steps

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

Use the common integral:

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

\frac{d}{d x} (9 x ln (x))

2^{- x}

x \to 400 - lim (x)

\int \frac{x ^{5}}{4 + x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (sin (2 x))

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