What is the integral of 1/((2x-3)(3x^2+x)) ?

The integral of 1/((2x-3)(3x^2+x)) is 2/33 ln|2x-3|-1/3 ln|x|+3/11 ln|3x+1|+C

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integral of 1/((2x-3)(3x^2+x))

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

\frac{2}{33} ln ∣ 2 x - 3 ∣ - \frac{1}{3} ln ∣ x ∣ + \frac{3}{11} ln ∣ 3 x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( 2 x - 3 ) ( 3 x ^{2} + x )} : \frac{4}{33 ( 2 x - 3 )} - \frac{1}{3 x} + \frac{9}{11 ( 3 x + 1 )}

= \int \frac{4}{33 ( 2 x - 3 )} - \frac{1}{3 x} + \frac{9}{11 ( 3 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{4}{33 ( 2 x - 3 )} d x - \int \frac{1}{3 x} d x + \int \frac{9}{11 ( 3 x + 1 )} d x

\int \frac{4}{33 ( 2 x - 3 )} d x = \frac{2}{33} ln ∣ 2 x - 3 ∣

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

\int \frac{9}{11 ( 3 x + 1 )} d x = \frac{3}{11} ln ∣ 3 x + 1 ∣

= \frac{2}{33} ln ∣ 2 x - 3 ∣ - \frac{1}{3} ln ∣ x ∣ + \frac{3}{11} ln ∣ 3 x + 1 ∣

Add a constant to the solution

= \frac{2}{33} ln ∣ 2 x - 3 ∣ - \frac{1}{3} ln ∣ x ∣ + \frac{3}{11} ln ∣ 3 x + 1 ∣ + C

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What is the integral of 1/((2x-3)(3x^2+x)) ?
The integral of 1/((2x-3)(3x^2+x)) is 2/33 ln|2x-3|-1/3 ln|x|+3/11 ln|3x+1|+C