What is the integral of 1/(x^3-4x^2) ?

The integral of 1/(x^3-4x^2) is -1/16 ln|x|+1/(4x)+1/16 ln|x-4|+C

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integral of 1/(x^3-4x^2)

\int \frac{1}{x ^{3} - 4 x ^{2}} d x

- \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{x ^{3} - 4 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} - 4 x ^{2}} : - \frac{1}{16 x} - \frac{1}{4 x ^{2}} + \frac{1}{16 ( x - 4 )}

= \int - \frac{1}{16 x} - \frac{1}{4 x ^{2}} + \frac{1}{16 ( x - 4 )} 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}{16 x} d x - \int \frac{1}{4 x ^{2}} d x + \int \frac{1}{16 ( x - 4 )} d x

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

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

\int \frac{1}{16 ( x - 4 )} d x = \frac{1}{16} ln ∣ x - 4 ∣

= - \frac{1}{16} ln ∣ x ∣ - (- \frac{1}{4 x}) + \frac{1}{16} ln ∣ x - 4 ∣

Simplify

= - \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣ + C

\frac{\partial}{\partial x} (\frac{9 x}{7 y + 8 z})

\frac{x d y}{d x} + 6 y = x^{3} - x

\int 8 x e^{2 x} d x

\frac{1}{m + k e ^{m}}

\int \frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}} d x

What is the integral of 1/(x^3-4x^2) ?
The integral of 1/(x^3-4x^2) is -1/16 ln|x|+1/(4x)+1/16 ln|x-4|+C