What is the integral of (10x^2-19x+6)/(x^3-3x^2+2x) ?

The integral of (10x^2-19x+6)/(x^3-3x^2+2x) is 3ln|x|+3ln|x-1|+4ln|x-2|+C

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integral of (10x^2-19x+6)/(x^3-3x^2+2x)

\int \frac{10 x ^{2} - 19 x + 6}{x ^{3} - 3 x ^{2} + 2 x} d x

3 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{10 x ^{2} - 19 x + 6}{x ^{3} - 3 x ^{2} + 2 x} d x

Take the partial fraction of

\frac{10 x ^{2} - 19 x + 6}{x ^{3} - 3 x ^{2} + 2 x} : \frac{3}{x} + \frac{3}{x - 1} + \frac{4}{x - 2}

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

= 3 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣

Add a constant to the solution

= 3 ln ∣ x ∣ + 3 ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + C

\int tan^{2} (x) \cdot sec^{2} (x) d x

\frac{d}{d x} ((x^{3} + 3) e^{x})

\frac{\partial}{\partial x} (A x + B y + C e^{x y})

\int_{0}^{5} \frac{1}{x ^{\frac{5}{2}}} d x

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

What is the integral of (10x^2-19x+6)/(x^3-3x^2+2x) ?
The integral of (10x^2-19x+6)/(x^3-3x^2+2x) is 3ln|x|+3ln|x-1|+4ln|x-2|+C