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

The integral of (2y+3)/((y-1)(y^3-1)) is

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integral of (2y+3)/((y-1)(y^3-1))

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

- ln ∣ y - 1 ∣ - \frac{5}{3 ( y - 1 )} + \frac{1}{3} (6 (\frac{1}{4} ln 4 y^{2} + 4 y + 4 - \frac{1}{2 3} arctan (\frac{2 y + 1}{3})) + \frac{2}{3} arctan (\frac{2 y + 1}{3})) + C

Solution steps

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

Take the partial fraction of

\frac{2 y + 3}{( y - 1 ) ( y ^{3} - 1 )} : - \frac{1}{y - 1} + \frac{5}{3 ( y - 1 ) ^{2}} + \frac{3 y + 1}{3 ( y ^{2} + y + 1 )}

= \int - \frac{1}{y - 1} + \frac{5}{3 ( y - 1 ) ^{2}} + \frac{3 y + 1}{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}{y - 1} d y + \int \frac{5}{3 ( y - 1 ) ^{2}} d y + \int \frac{3 y + 1}{3 ( y ^{2} + y + 1 )} d y

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

\int \frac{5}{3 ( y - 1 ) ^{2}} d y = - \frac{5}{3 ( y - 1 )}

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

= - ln ∣ y - 1 ∣ - \frac{5}{3 ( y - 1 )} + \frac{1}{3} (6 (\frac{1}{4} ln 4 y^{2} + 4 y + 4 - \frac{1}{2 3} arctan (\frac{2 y + 1}{3})) + \frac{2}{3} arctan (\frac{2 y + 1}{3}))

Add a constant to the solution

= - ln ∣ y - 1 ∣ - \frac{5}{3 ( y - 1 )} + \frac{1}{3} (6 (\frac{1}{4} ln 4 y^{2} + 4 y + 4 - \frac{1}{2 3} arctan (\frac{2 y + 1}{3})) + \frac{2}{3} arctan (\frac{2 y + 1}{3})) + C

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What is the integral of (2y+3)/((y-1)(y^3-1)) ?
The integral of (2y+3)/((y-1)(y^3-1)) is