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

The integral of 1/(6x^3-7x^2-3x) is -1/3 ln|x|+3/11 ln|3x+1|+2/33 ln|2x-3|+C

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

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{6 x ^{3} - 7 x ^{2} - 3 x} : - \frac{1}{3 x} + \frac{9}{11 ( 3 x + 1 )} + \frac{4}{33 ( 2 x - 3 )}

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

\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 ∣

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

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

Add a constant to the solution

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

\frac{\partial}{\partial v} (sin (u) cos (v))

y^{^{'}} = 5 y - x y

y^{^{''}} + 8 y^{^{'}} + 25 y = - 208 \cdot sin (3 x)

x \to 3 lim (3 - ∣ x - 3 ∣)

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

What is the integral of 1/(6x^3-7x^2-3x) ?
The integral of 1/(6x^3-7x^2-3x) is -1/3 ln|x|+3/11 ln|3x+1|+2/33 ln|2x-3|+C