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

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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{s ^{4} - 2 s ^{3} + s ^{2}} : \frac{2}{s} + \frac{1}{s ^{2}} - \frac{2}{s - 1} + \frac{1}{( s - 1 ) ^{2}}

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

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

\int \frac{2}{s} d s = 2 ln ∣ s ∣

\int \frac{1}{s ^{2}} d s = - \frac{1}{s}

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

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

= 2 ln ∣ s ∣ - \frac{1}{s} - 2 ln ∣ s - 1 ∣ - \frac{1}{s - 1}

Add a constant to the solution

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

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