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

\int_{0}^{5} \frac{4 t}{( t - 6 ) ^{2}} d t

\int \frac{( 4 u ^{2} - 8 )}{2 u ^{2} + 1} d u

\frac{d}{d x} (\frac{x ^{2}}{4 - x ^{2}})

- \frac{24}{x ^{4}}

\int p (p + 3)^{6} d p

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