What is the integral from 0 to 1 of (x-6)/(x^2-5x+6) ?

The integral from 0 to 1 of (x-6)/(x^2-5x+6) is (-ln(3)-5ln(3/4))/2-6(2ln(2)-ln(3))

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integral from 0 to 1 of (x-6)/(x^2-5x+6)

\int_{0}^{1} \frac{x - 6}{x ^{2} - 5 x + 6} d x

\frac{- ln ( 3 ) - 5 ln ( \frac{3}{4} )}{2} - 6 (2 ln (2) - ln (3))

Decimal Notation

- 1.55619 \dots

Solution steps

\int_{0}^{1} \frac{x - 6}{x ^{2} - 5 x + 6} d x

Expand

\frac{x - 6}{x ^{2} - 5 x + 6} : \frac{x}{x ^{2} - 5 x + 6} - \frac{6}{x ^{2} - 5 x + 6}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} \frac{x}{x ^{2} - 5 x + 6} d x - \int_{0}^{1} \frac{6}{x ^{2} - 5 x + 6} d x

\int_{0}^{1} \frac{x}{x ^{2} - 5 x + 6} d x = \frac{- ln ( 3 ) - 5 ln ( \frac{3}{4} )}{2}

\int_{0}^{1} \frac{6}{x ^{2} - 5 x + 6} d x = 6 (2 ln (2) - ln (3))

= \frac{- ln ( 3 ) - 5 ln ( \frac{3}{4} )}{2} - 6 (2 ln (2) - ln (3))

\frac{d y}{d x} - e^{8 x + y} = 0

\int x \cdot sin (2 a x) d x

cos (e^{x^{2 c o s (x)}})

\frac{d}{d x} (5 e^{- x} - 3 e^{- 5 x})

f (x) = (x^{2} + 1) \cdot e^{x - 1}

What is the integral from 0 to 1 of (x-6)/(x^2-5x+6) ?
The integral from 0 to 1 of (x-6)/(x^2-5x+6) is (-ln(3)-5ln(3/4))/2-6(2ln(2)-ln(3))