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

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

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

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

- \frac{1}{9} ln ∣ x ∣ + \frac{1}{3 x} + \frac{1}{9} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

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

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

= - \frac{1}{9} ln ∣ x ∣ - (- \frac{1}{3 x}) + \frac{1}{9} ln ∣ x - 3 ∣

Simplify

= - \frac{1}{9} ln ∣ x ∣ + \frac{1}{3 x} + \frac{1}{9} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{9} ln ∣ x ∣ + \frac{1}{3 x} + \frac{1}{9} ln ∣ x - 3 ∣ + C

\frac{d}{d x} (x^{2} - 3 x + 12)

f (x) = 2 sin (x), at x = \frac{3 π}{4}

y^{^{'}} = sin (5 x), y (0) = 2

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 3 x yz)

(x + 1) \frac{d y}{d x} + (x + 2) y = 8 x e^{- x}

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