What is the integral of (x^2-4x-3)/(3x^3+17x^2-6x) ?

The integral of (x^2-4x-3)/(3x^3+17x^2-6x) is 1/2 ln|x|-2/3 ln|3x-1|+1/2 ln|x+6|+C

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integral of (x^2-4x-3)/(3x^3+17x^2-6x)

\int \frac{x ^{2} - 4 x - 3}{3 x ^{3} + 17 x ^{2} - 6 x} d x

\frac{1}{2} ln ∣ x ∣ - \frac{2}{3} ln ∣ 3 x - 1 ∣ + \frac{1}{2} ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{x ^{2} - 4 x - 3}{3 x ^{3} + 17 x ^{2} - 6 x} d x

Take the partial fraction of

\frac{x ^{2} - 4 x - 3}{3 x ^{3} + 17 x ^{2} - 6 x} : \frac{1}{2 x} - \frac{2}{3 x - 1} + \frac{1}{2 ( x + 6 )}

= \int \frac{1}{2 x} - \frac{2}{3 x - 1} + \frac{1}{2 ( x + 6 )} 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{1}{2 x} d x - \int \frac{2}{3 x - 1} d x + \int \frac{1}{2 ( x + 6 )} d x

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

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

\int \frac{1}{2 ( x + 6 )} d x = \frac{1}{2} ln ∣ x + 6 ∣

= \frac{1}{2} ln ∣ x ∣ - \frac{2}{3} ln ∣ 3 x - 1 ∣ + \frac{1}{2} ln ∣ x + 6 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ - \frac{2}{3} ln ∣ 3 x - 1 ∣ + \frac{1}{2} ln ∣ x + 6 ∣ + C

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What is the integral of (x^2-4x-3)/(3x^3+17x^2-6x) ?
The integral of (x^2-4x-3)/(3x^3+17x^2-6x) is 1/2 ln|x|-2/3 ln|3x-1|+1/2 ln|x+6|+C