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

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