What is the integral of ((x-4))/(x^2-5x+6) ?

The integral of ((x-4))/(x^2-5x+6) is -ln|x-3|+2ln|x-2|+C

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integral of ((x-4))/(x^2-5x+6)

\int \frac{( x - 4 )}{x ^{2} - 5 x + 6} d x

- ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x - 4}{x ^{2} - 5 x + 6} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 3 u + 2} d u

= \int \frac{u}{u ^{2} + 3 u + 2} d u

Take the partial fraction of

\frac{u}{u ^{2} + 3 u + 2} : - \frac{1}{u + 1} + \frac{2}{u + 2}

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

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

\int \frac{2}{u + 2} d u = 2 ln ∣ u + 2 ∣

= - ln ∣ u + 1 ∣ + 2 ln ∣ u + 2 ∣

Substitute back

u = x - 4

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

Simplify

= - ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x - 3 ∣ + 2 ln ∣ x - 2 ∣ + C

\int x^{4} e^{1 - x^{5}} d x

2 x^{3} + 3 x^{2} - 12 x + 1

\frac{d}{d y} (y e^{x y} cos (2 x) - 2 e^{x y} sin (2 x) + 2 x)

n = 1 \sum \infty \frac{( n ! ) ^{2}}{1}

x \to a lim (3 x^{2} - 6 x)

What is the integral of ((x-4))/(x^2-5x+6) ?
The integral of ((x-4))/(x^2-5x+6) is -ln|x-3|+2ln|x-2|+C