What is the integral of (y^2+6)/(y(y^2-y-6)) ?

The integral of (y^2+6)/(y(y^2-y-6)) is

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integral of (y^2+6)/(y(y^2-y-6))

\int \frac{y ^{2} + 6}{y ( y ^{2} - y - 6 )} d y

2 (\frac{1}{4} ln 4 y^{2} - 4 y - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} y - ln - \frac{6}{5} + \frac{2}{5} y)) - ln ∣ y ∣ + \frac{3}{5} ln ∣ y + 2 ∣ + \frac{2}{5} ln ∣ y - 3 ∣ + C

Solution steps

\int \frac{y ^{2} + 6}{y ( y ^{2} - y - 6 )} d y

Expand

\frac{y ^{2} + 6}{y ( y ^{2} - y - 6 )} : \frac{y ^{2}}{y ( y ^{2} - y - 6 )} + \frac{6}{y ( y ^{2} - y - 6 )}

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{y ^{2}}{y ( y ^{2} - y - 6 )} d y + \int \frac{6}{y ( y ^{2} - y - 6 )} d y

\int \frac{y ^{2}}{y ( y ^{2} - y - 6 )} d y = 2 (\frac{1}{4} ln 4 y^{2} - 4 y - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} y - ln - \frac{6}{5} + \frac{2}{5} y))

\int \frac{6}{y ( y ^{2} - y - 6 )} d y = - ln ∣ y ∣ + \frac{3}{5} ln ∣ y + 2 ∣ + \frac{2}{5} ln ∣ y - 3 ∣

= 2 (\frac{1}{4} ln 4 y^{2} - 4 y - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} y - ln - \frac{6}{5} + \frac{2}{5} y)) - ln ∣ y ∣ + \frac{3}{5} ln ∣ y + 2 ∣ + \frac{2}{5} ln ∣ y - 3 ∣

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 y^{2} - 4 y - 24 - \frac{1}{20} (ln \frac{4}{5} + \frac{2}{5} y - ln - \frac{6}{5} + \frac{2}{5} y)) - ln ∣ y ∣ + \frac{3}{5} ln ∣ y + 2 ∣ + \frac{2}{5} ln ∣ y - 3 ∣ + C

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What is the integral of (y^2+6)/(y(y^2-y-6)) ?
The integral of (y^2+6)/(y(y^2-y-6)) is