What is the integral of (2x^2-12x+4)/(x^3-4x^2) ?

The integral of (2x^2-12x+4)/(x^3-4x^2) is 11/4 ln|x|+1/x-3/4 ln|x-4|+C

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integral of (2x^2-12x+4)/(x^3-4x^2)

\int \frac{2 x ^{2} - 12 x + 4}{x ^{3} - 4 x ^{2}} d x

\frac{11}{4} ln ∣ x ∣ + \frac{1}{x} - \frac{3}{4} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{2 x ^{2} - 12 x + 4}{x ^{3} - 4 x ^{2}} d x

Take the partial fraction of

\frac{2 x ^{2} - 12 x + 4}{x ^{3} - 4 x ^{2}} : \frac{11}{4 x} - \frac{1}{x ^{2}} - \frac{3}{4 ( x - 4 )}

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

\int \frac{11}{4 x} d x = \frac{11}{4} ln ∣ x ∣

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

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

= \frac{11}{4} ln ∣ x ∣ - (- \frac{1}{x}) - \frac{3}{4} ln ∣ x - 4 ∣

Simplify

= \frac{11}{4} ln ∣ x ∣ + \frac{1}{x} - \frac{3}{4} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{11}{4} ln ∣ x ∣ + \frac{1}{x} - \frac{3}{4} ln ∣ x - 4 ∣ + C

What is the integral of (2x^2-12x+4)/(x^3-4x^2) ?
The integral of (2x^2-12x+4)/(x^3-4x^2) is 11/4 ln|x|+1/x-3/4 ln|x-4|+C