What is the integral of (1-6v)/(1-5v+6v^2) ?

The integral of (1-6v)/(1-5v+6v^2) is ln|3v-1|-2ln|2v-1|+C

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integral of (1-6v)/(1-5v+6v^2)

\int \frac{1 - 6 v}{1 - 5 v + 6 v ^{2}} d v

ln ∣ 3 v - 1 ∣ - 2 ln ∣ 2 v - 1 ∣ + C

Solution steps

\int \frac{1 - 6 v}{1 - 5 v + 6 v ^{2}} d v

Take the partial fraction of

\frac{1 - 6 v}{1 - 5 v + 6 v ^{2}} : \frac{3}{3 v - 1} - \frac{4}{2 v - 1}

= \int \frac{3}{3 v - 1} - \frac{4}{2 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{3}{3 v - 1} d v - \int \frac{4}{2 v - 1} d v

\int \frac{3}{3 v - 1} d v = ln ∣ 3 v - 1 ∣

\int \frac{4}{2 v - 1} d v = 2 ln ∣ 2 v - 1 ∣

= ln ∣ 3 v - 1 ∣ - 2 ln ∣ 2 v - 1 ∣

Add a constant to the solution

= ln ∣ 3 v - 1 ∣ - 2 ln ∣ 2 v - 1 ∣ + C

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What is the integral of (1-6v)/(1-5v+6v^2) ?
The integral of (1-6v)/(1-5v+6v^2) is ln|3v-1|-2ln|2v-1|+C