What is the integral from 0 to 1 of (x-8)/(x^2-5x+6) ?

The integral from 0 to 1 of (x-8)/(x^2-5x+6) is (-ln(3)+11ln(3/4))/2

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integral from 0 to 1 of (x-8)/(x^2-5x+6)

\int_{0}^{1} \frac{x - 8}{x ^{2} - 5 x + 6} d x

\frac{- ln ( 3 ) + 11 ln ( \frac{3}{4} )}{2}

Decimal Notation

- 2.13155 \dots

Solution steps

\int_{0}^{1} \frac{x - 8}{x ^{2} - 5 x + 6} d x

Apply u-substitution

: \int_{- 8}^{- 7} \frac{u}{u ^{2} + 11 u + 30} d u

= \int_{- 8}^{- 7} \frac{u}{u ^{2} + 11 u + 30} d u

Complete the square

u^{2} + 11 u + 30 : (u + \frac{11}{2})^{2} - \frac{1}{4}

= \int_{- 8}^{- 7} \frac{u}{( u + \frac{11}{2} ) ^{2} - \frac{1}{4}} d u

Apply u-substitution

: \int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 ( 2 v - 11 )}{4 v ^{2} - 1} d v

= \int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 ( 2 v - 11 )}{4 v ^{2} - 1} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 v - 11}{4 v ^{2} - 1} d v

Expand

\frac{2 v - 11}{4 v ^{2} - 1} : \frac{2 v}{4 v ^{2} - 1} - \frac{11}{4 v ^{2} - 1}

Apply the Sum Rule:

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

= 2 (\int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 v}{4 v ^{2} - 1} d v - \int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{11}{4 v ^{2} - 1} d v)

\int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 v}{4 v ^{2} - 1} d v = - \frac{1}{4} ln (3)

\int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{11}{4 v ^{2} - 1} d v = - \frac{11 ln ( \frac{3}{4} )}{4}

= 2 (- \frac{1}{4} ln (3) - (- \frac{11 ln ( \frac{3}{4} )}{4}))

Simplify

2 (- \frac{1}{4} ln (3) - (- \frac{11 ln ( \frac{3}{4} )}{4})) : \frac{- ln ( 3 ) + 11 ln ( \frac{3}{4} )}{2}

= \frac{- ln ( 3 ) + 11 ln ( \frac{3}{4} )}{2}

\frac{d}{d x} (\frac{x ^{2}}{ln ( x )})

\frac{d}{d x} (x^{\frac{3}{5}} x^{3} + 2 x - 6)

\frac{d}{d x} (2 x^{2} - 9 x + 10)

\frac{2 s - 44}{7 s ^{2} - 14 s + 70}

\frac{d y}{d x} - sin (x + y) = 0

What is the integral from 0 to 1 of (x-8)/(x^2-5x+6) ?
The integral from 0 to 1 of (x-8)/(x^2-5x+6) is (-ln(3)+11ln(3/4))/2