What is the integral of (13x-4)/(6x^2-x-2) ?

The integral of (13x-4)/(6x^2-x-2) is 3/2 ln|2x+1|+2/3 ln|3x-2|+C

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integral of (13x-4)/(6x^2-x-2)

\int \frac{13 x - 4}{6 x ^{2} - x - 2} d x

\frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ 3 x - 2 ∣ + C

Solution steps

\int \frac{13 x - 4}{6 x ^{2} - x - 2} d x

Take the partial fraction of

\frac{13 x - 4}{6 x ^{2} - x - 2} : \frac{3}{2 x + 1} + \frac{2}{3 x - 2}

= \int \frac{3}{2 x + 1} + \frac{2}{3 x - 2} d x

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}{2 x + 1} d x + \int \frac{2}{3 x - 2} d x

\int \frac{3}{2 x + 1} d x = \frac{3}{2} ln ∣ 2 x + 1 ∣

\int \frac{2}{3 x - 2} d x = \frac{2}{3} ln ∣ 3 x - 2 ∣

= \frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ 3 x - 2 ∣

Add a constant to the solution

= \frac{3}{2} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ 3 x - 2 ∣ + C

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What is the integral of (13x-4)/(6x^2-x-2) ?
The integral of (13x-4)/(6x^2-x-2) is 3/2 ln|2x+1|+2/3 ln|3x-2|+C