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

x^{2} y^{^{''}} = (y^{^{'}})^{2}

\frac{\partial}{\partial x} (3 y^{3} - 4 y x^{2})

\int_{0}^{7} f (x) d x

θ \to 0 lim (\frac{sin ( 4 θ )}{θ})

f (x) = 8 x^{2} + 4 e^{- 0.5 x}

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