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

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

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integral of 1/(y^2-y)

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

- ln ∣ y ∣ + ln ∣ y - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{y ^{2} - y} : - \frac{1}{y} + \frac{1}{y - 1}

= \int - \frac{1}{y} + \frac{1}{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} d y + \int \frac{1}{y - 1} d y

\int \frac{1}{y} d y = ln ∣ y ∣

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

= - ln ∣ y ∣ + ln ∣ y - 1 ∣

Add a constant to the solution

= - ln ∣ y ∣ + ln ∣ y - 1 ∣ + C

f (x) = x^{2} + 2

\frac{\partial}{\partial x} (x y e^{x + y})

x \to 1 lim (\frac{x ^{2} + 7 x}{x})

t^{3}

\int_{- \infty}^{0} \frac{1}{3 - 6 x} d x

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