What is the integral of (2x^2-3x+1)/(x^3-6x^2+8x) ?

The integral of (2x^2-3x+1)/(x^3-6x^2+8x) is 1/8 ln|x|-3/4 ln|x-2|+21/8 ln|x-4|+C

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integral of (2x^2-3x+1)/(x^3-6x^2+8x)

\int \frac{2 x ^{2} - 3 x + 1}{x ^{3} - 6 x ^{2} + 8 x} d x

\frac{1}{8} ln ∣ x ∣ - \frac{3}{4} ln ∣ x - 2 ∣ + \frac{21}{8} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{2 x ^{2} - 3 x + 1}{x ^{3} - 6 x ^{2} + 8 x} d x

Take the partial fraction of

\frac{2 x ^{2} - 3 x + 1}{x ^{3} - 6 x ^{2} + 8 x} : \frac{1}{8 x} - \frac{3}{4 ( x - 2 )} + \frac{21}{8 ( x - 4 )}

= \int \frac{1}{8 x} - \frac{3}{4 ( x - 2 )} + \frac{21}{8 ( 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}{8 x} d x - \int \frac{3}{4 ( x - 2 )} d x + \int \frac{21}{8 ( x - 4 )} d x

\int \frac{1}{8 x} d x = \frac{1}{8} ln ∣ x ∣

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

\int \frac{21}{8 ( x - 4 )} d x = \frac{21}{8} ln ∣ x - 4 ∣

= \frac{1}{8} ln ∣ x ∣ - \frac{3}{4} ln ∣ x - 2 ∣ + \frac{21}{8} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ x ∣ - \frac{3}{4} ln ∣ x - 2 ∣ + \frac{21}{8} ln ∣ x - 4 ∣ + C

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What is the integral of (2x^2-3x+1)/(x^3-6x^2+8x) ?
The integral of (2x^2-3x+1)/(x^3-6x^2+8x) is 1/8 ln|x|-3/4 ln|x-2|+21/8 ln|x-4|+C