What is the integral of 3/(w^3-w) ?

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

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integral of 3/(w^3-w)

\int \frac{3}{w ^{3} - w} d w

3 (- ln ∣ w ∣ + \frac{1}{2} ln ∣ w + 1 ∣ + \frac{1}{2} ln ∣ w - 1 ∣) + C

Solution steps

\int \frac{3}{w ^{3} - w} d w

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{w ^{3} - w} : - \frac{1}{w} + \frac{1}{2 ( w + 1 )} + \frac{1}{2 ( w - 1 )}

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

Apply the Sum Rule:

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

= 3 (- \int \frac{1}{w} d w + \int \frac{1}{2 ( w + 1 )} d w + \int \frac{1}{2 ( w - 1 )} d w)

\int \frac{1}{w} d w = ln ∣ w ∣

\int \frac{1}{2 ( w + 1 )} d w = \frac{1}{2} ln ∣ w + 1 ∣

\int \frac{1}{2 ( w - 1 )} d w = \frac{1}{2} ln ∣ w - 1 ∣

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

Add a constant to the solution

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

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