What is the integral of 1/(6x^3-7x^2+2x) ?

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

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integral of 1/(6x^3-7x^2+2x)

\int \frac{1}{6 x ^{3} - 7 x ^{2} + 2 x} d x

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

Solution steps

\int \frac{1}{6 x ^{3} - 7 x ^{2} + 2 x} d x

Take the partial fraction of

\frac{1}{6 x ^{3} - 7 x ^{2} + 2 x} : \frac{1}{2 x} - \frac{4}{2 x - 1} + \frac{9}{2 ( 3 x - 2 )}

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

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

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

\int \frac{9}{2 ( 3 x - 2 )} d x = \frac{3}{2} ln ∣ 3 x - 2 ∣

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

Add a constant to the solution

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

\int \frac{x - 4}{x ( x - 2 ) ( x - 5 )} d x

\int \frac{40}{S ^{2}} d S

\int \frac{2}{( 5 - 3 x ) ^{9}} d x

\int ln (2 x - 3) d x

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

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