What is the integral of ((x^2-x)/(x^3-6x^2+11x-6)) ?

The integral of ((x^2-x)/(x^3-6x^2+11x-6)) is -2ln|x-2|+3ln|x-3|+C

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integral of ((x^2-x)/(x^3-6x^2+11x-6))

\int (\frac{x ^{2} - x}{x ^{3} - 6 x ^{2} + 11 x - 6}) d x

- 2 ln ∣ x - 2 ∣ + 3 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{2} - x}{x ^{3} - 6 x ^{2} + 11 x - 6} d x

Simplify

\frac{x ^{2} - x}{x ^{3} - 6 x ^{2} + 11 x - 6} : \frac{x}{( x - 2 ) ( x - 3 )}

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

Apply u-substitution

: \int \frac{u + 2}{u ( u - 1 )} d u

= \int \frac{u + 2}{u ( u - 1 )} d u

Take the partial fraction of

\frac{u + 2}{u ( u - 1 )} : - \frac{2}{u} + \frac{3}{u - 1}

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

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{2}{u} d u + \int \frac{3}{u - 1} d u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= - 2 ln ∣ u ∣ + 3 ln ∣ u - 1 ∣

Substitute back

u = x - 2

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

Simplify

= - 2 ln ∣ x - 2 ∣ + 3 ln ∣ x - 3 ∣

Add a constant to the solution

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

\frac{d}{d x} (e^{g (x) (x^{3}) - 2 x})

\frac{d}{d x} (\frac{x}{x + y})

y = e^{5 x}

x^{3} - 8 x^{2}

\int_{- 1}^{1} (1 - x^{2}) d x

What is the integral of ((x^2-x)/(x^3-6x^2+11x-6)) ?
The integral of ((x^2-x)/(x^3-6x^2+11x-6)) is -2ln|x-2|+3ln|x-3|+C