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

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

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

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

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

Solution steps

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

Use the common integral:

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

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

Add a constant to the solution

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

8 csc (5 x^{4} - 8 x + 6)

n = 0 \sum \infty - \frac{1}{3 ^{n}}

x \to 0 lim (\frac{1}{x ^{2} + 1})

\frac{d}{d x} (sec^{2} (x) - 4)

\frac{d y}{d x} = \frac{1}{x ^{2}}

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