What is the integral of (s+1)/(s^3+s^2-6s) ?

The integral of (s+1)/(s^3+s^2-6s) is -1/6 ln|s|+3/10 ln|s-2|-2/15 ln|s+3|+C

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integral of (s+1)/(s^3+s^2-6s)

\int \frac{s + 1}{s ^{3} + s ^{2} - 6 s} d s

- \frac{1}{6} ln ∣ s ∣ + \frac{3}{10} ln ∣ s - 2 ∣ - \frac{2}{15} ln ∣ s + 3 ∣ + C

Solution steps

\int \frac{s + 1}{s ^{3} + s ^{2} - 6 s} d s

Take the partial fraction of

\frac{s + 1}{s ^{3} + s ^{2} - 6 s} : - \frac{1}{6 s} + \frac{3}{10 ( s - 2 )} - \frac{2}{15 ( s + 3 )}

= \int - \frac{1}{6 s} + \frac{3}{10 ( s - 2 )} - \frac{2}{15 ( s + 3 )} 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{1}{6 s} d s + \int \frac{3}{10 ( s - 2 )} d s - \int \frac{2}{15 ( s + 3 )} d s

\int \frac{1}{6 s} d s = \frac{1}{6} ln ∣ s ∣

\int \frac{3}{10 ( s - 2 )} d s = \frac{3}{10} ln ∣ s - 2 ∣

\int \frac{2}{15 ( s + 3 )} d s = \frac{2}{15} ln ∣ s + 3 ∣

= - \frac{1}{6} ln ∣ s ∣ + \frac{3}{10} ln ∣ s - 2 ∣ - \frac{2}{15} ln ∣ s + 3 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ s ∣ + \frac{3}{10} ln ∣ s - 2 ∣ - \frac{2}{15} ln ∣ s + 3 ∣ + C

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What is the integral of (s+1)/(s^3+s^2-6s) ?
The integral of (s+1)/(s^3+s^2-6s) is -1/6 ln|s|+3/10 ln|s-2|-2/15 ln|s+3|+C