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

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

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integral of 1/(v^2-v)

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

- ln ∣ v ∣ + ln ∣ v - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

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}{v} d v + \int \frac{1}{v - 1} d v

\int \frac{1}{v} d v = ln ∣ v ∣

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

= - ln ∣ v ∣ + ln ∣ v - 1 ∣

Add a constant to the solution

= - ln ∣ v ∣ + ln ∣ v - 1 ∣ + C

f (θ) = 3 sin (θ)

\frac{\partial}{\partial y} (5 x y^{2} + x^{2})

t^{2} y^{^{''}} + t y^{^{'}} + 64 y = 0

\frac{d}{d v} (\frac{m}{v})

1 - 2 x

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