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

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

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

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

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

Solution steps

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

Factor

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d v = v

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

Simplify

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

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

Add a constant to the solution

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

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