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

The integral of 1/(s^3-1) is 1/3 ln|s-1|-1/6 ln|4s^2+4s+4|-1/(sqrt(3))arctan(1/(sqrt(3))(2s+1))+C

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integral of 1/(s^3-1)

\int \frac{1}{s ^{3} - 1} d s

\frac{1}{3} ln ∣ s - 1 ∣ - \frac{1}{6} ln 4 s^{2} + 4 s + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 s + 1)) + C

Solution steps

\int \frac{1}{s ^{3} - 1} d s

Take the partial fraction of

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

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

\int \frac{1}{3 ( s - 1 )} d s = \frac{1}{3} ln ∣ s - 1 ∣

\int \frac{- s - 2}{3 ( s ^{2} + s + 1 )} d s = \frac{1}{6 3} (- 3 ln 4 s^{2} + 4 s + 4 - 6 arctan (\frac{2 s + 1}{3}))

= \frac{1}{3} ln ∣ s - 1 ∣ + \frac{1}{6 3} (- 3 ln 4 s^{2} + 4 s + 4 - 6 arctan (\frac{2 s + 1}{3}))

Simplify

\frac{1}{3} ln ∣ s - 1 ∣ + \frac{1}{6 3} (- 3 ln 4 s^{2} + 4 s + 4 - 6 arctan (\frac{2 s + 1}{3})) : \frac{1}{3} ln ∣ s - 1 ∣ - \frac{1}{6} ln 4 s^{2} + 4 s + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 s + 1))

= \frac{1}{3} ln ∣ s - 1 ∣ - \frac{1}{6} ln 4 s^{2} + 4 s + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 s + 1))

Add a constant to the solution

= \frac{1}{3} ln ∣ s - 1 ∣ - \frac{1}{6} ln 4 s^{2} + 4 s + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 s + 1)) + C

\frac{d}{d x} (x arctan (2 x) - \frac{1}{4} ln (1 + 4 x^{2}))

9 x^{\frac{3}{5}} - 2 x^{\frac{6}{7}}, at x = 3

x \to - 4 lim (- (x^{2} + 10 x + 8))

f (x) = \frac{3}{( 2 x ) ^{2}}

f (x) = tan (x), (\frac{3 π}{4}, - 1)

What is the integral of 1/(s^3-1) ?
The integral of 1/(s^3-1) is 1/3 ln|s-1|-1/6 ln|4s^2+4s+4|-1/(sqrt(3))arctan(1/(sqrt(3))(2s+1))+C