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

The integral of (x+2)/(6x^3+x^2-x) is -2ln|x|+7/5 ln|3x-1|+3/5 ln|2x+1|+C

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

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

- 2 ln ∣ x ∣ + \frac{7}{5} ln ∣ 3 x - 1 ∣ + \frac{3}{5} ln ∣ 2 x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x + 2}{6 x ^{3} + x ^{2} - x} : - \frac{2}{x} + \frac{21}{5 ( 3 x - 1 )} + \frac{6}{5 ( 2 x + 1 )}

= \int - \frac{2}{x} + \frac{21}{5 ( 3 x - 1 )} + \frac{6}{5 ( 2 x + 1 )} 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{2}{x} d x + \int \frac{21}{5 ( 3 x - 1 )} d x + \int \frac{6}{5 ( 2 x + 1 )} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{21}{5 ( 3 x - 1 )} d x = \frac{7}{5} ln ∣ 3 x - 1 ∣

\int \frac{6}{5 ( 2 x + 1 )} d x = \frac{3}{5} ln ∣ 2 x + 1 ∣

= - 2 ln ∣ x ∣ + \frac{7}{5} ln ∣ 3 x - 1 ∣ + \frac{3}{5} ln ∣ 2 x + 1 ∣

Add a constant to the solution

= - 2 ln ∣ x ∣ + \frac{7}{5} ln ∣ 3 x - 1 ∣ + \frac{3}{5} ln ∣ 2 x + 1 ∣ + C

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What is the integral of (x+2)/(6x^3+x^2-x) ?
The integral of (x+2)/(6x^3+x^2-x) is -2ln|x|+7/5 ln|3x-1|+3/5 ln|2x+1|+C