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

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

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

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

- \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ y - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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

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

= - \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ y - 3 ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ y ∣ + \frac{1}{3} ln ∣ y - 3 ∣ + C

y = - x + 4, y = \frac{x ^{2}}{3}

\frac{d}{d x} (csc^{2} (x) - 4)

\frac{4}{x ^{2} + x}

e^{- x}, 0

\frac{d}{d x} ((2 x + 5) (3 x^{2} - 4 x + 1))

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