What is the integral of 1/(9x^3-2x^2) ?

The integral of 1/(9x^3-2x^2) is -9/4 ln|x|+1/(2x)+9/4 ln|9x-2|+C

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integral of 1/(9x^3-2x^2)

\int \frac{1}{9 x ^{3} - 2 x ^{2}} d x

- \frac{9}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{9}{4} ln ∣ 9 x - 2 ∣ + C

Solution steps

\int \frac{1}{9 x ^{3} - 2 x ^{2}} d x

Take the partial fraction of

\frac{1}{9 x ^{3} - 2 x ^{2}} : - \frac{9}{4 x} - \frac{1}{2 x ^{2}} + \frac{81}{4 ( 9 x - 2 )}

= \int - \frac{9}{4 x} - \frac{1}{2 x ^{2}} + \frac{81}{4 ( 9 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{9}{4 x} d x - \int \frac{1}{2 x ^{2}} d x + \int \frac{81}{4 ( 9 x - 2 )} d x

\int \frac{9}{4 x} d x = \frac{9}{4} ln ∣ x ∣

\int \frac{1}{2 x ^{2}} d x = - \frac{1}{2 x}

\int \frac{81}{4 ( 9 x - 2 )} d x = \frac{9}{4} ln ∣ 9 x - 2 ∣

= - \frac{9}{4} ln ∣ x ∣ - (- \frac{1}{2 x}) + \frac{9}{4} ln ∣ 9 x - 2 ∣

Simplify

= - \frac{9}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{9}{4} ln ∣ 9 x - 2 ∣

Add a constant to the solution

= - \frac{9}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{9}{4} ln ∣ 9 x - 2 ∣ + C

y = x^{2} - 3 x + 3, (0, 3)

\frac{d}{d x} (\frac{x ^{2} - 1}{sin ( x )})

f (x) = \frac{A}{x ^{7}} + B e^{7}

\frac{d}{d x} (cos (2))

\int_{1}^{3} (x^{2} + 1)^{3} d x

What is the integral of 1/(9x^3-2x^2) ?
The integral of 1/(9x^3-2x^2) is -9/4 ln|x|+1/(2x)+9/4 ln|9x-2|+C