What is the integral of (v^2)/(v^6-1) ?

The integral of (v^2)/(v^6-1) is -1/6 (ln|v^3+1|-ln|v^3-1|)+C

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integral of (v^2)/(v^6-1)

\int \frac{v ^{2}}{v ^{6} - 1} d v

- \frac{1}{6} (ln v^{3} + 1 - ln v^{3} - 1) + C

Solution steps

\int \frac{v ^{2}}{v ^{6} - 1} d v

Apply u-substitution

: \int \frac{1}{3 ( u ^{2} - 1 )} d u

= \int \frac{1}{3 ( u ^{2} - 1 )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{3} \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{3} (- \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{1}{3} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = v^{3}

= \frac{1}{3} (- (\frac{ln v ^{3} + 1}{2} - \frac{ln v ^{3} - 1}{2}))

Simplify

\frac{1}{3} (- (\frac{ln v ^{3} + 1}{2} - \frac{ln v ^{3} - 1}{2})) : - \frac{1}{6} (ln v^{3} + 1 - ln v^{3} - 1)

= - \frac{1}{6} (ln v^{3} + 1 - ln v^{3} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln v^{3} + 1 - ln v^{3} - 1) + C

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What is the integral of (v^2)/(v^6-1) ?
The integral of (v^2)/(v^6-1) is -1/6 (ln|v^3+1|-ln|v^3-1|)+C