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

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

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integral of 1/(y^3-1)

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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