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

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

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integral of 1/(v^2-1)

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

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

Solution steps

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

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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})

Simplify

- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}) : - \frac{1}{2} ln ∣ v + 1 ∣ + \frac{1}{2} ln ∣ v - 1 ∣

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

Add a constant to the solution

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

\frac{d}{d x} (ln (x - 1))

\int \frac{1}{x ( 1 - 2 x )} d x

y = \frac{7}{ln ( x )}

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

x \to 9 - lim (\frac{1}{x - 9})

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