What is the integral of (w^2+4w-1)/(w^3-w) ?

The integral of (w^2+4w-1)/(w^3-w) is ln|w|-2ln|w+1|+2ln|w-1|+C

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integral of (w^2+4w-1)/(w^3-w)

\int \frac{w ^{2} + 4 w - 1}{w ^{3} - w} d w

ln ∣ w ∣ - 2 ln ∣ w + 1 ∣ + 2 ln ∣ w - 1 ∣ + C

Solution steps

\int \frac{w ^{2} + 4 w - 1}{w ^{3} - w} d w

Take the partial fraction of

\frac{w ^{2} + 4 w - 1}{w ^{3} - w} : \frac{1}{w} - \frac{2}{w + 1} + \frac{2}{w - 1}

= \int \frac{1}{w} - \frac{2}{w + 1} + \frac{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

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

\int \frac{1}{w} d w = ln ∣ w ∣

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

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

= ln ∣ w ∣ - 2 ln ∣ w + 1 ∣ + 2 ln ∣ w - 1 ∣

Add a constant to the solution

= ln ∣ w ∣ - 2 ln ∣ w + 1 ∣ + 2 ln ∣ w - 1 ∣ + C

What is the integral of (w^2+4w-1)/(w^3-w) ?
The integral of (w^2+4w-1)/(w^3-w) is ln|w|-2ln|w+1|+2ln|w-1|+C