What is the integral from 0 to 1 of (x-9)/(x^2-5x+6) ?

The integral from 0 to 1 of (x-9)/(x^2-5x+6) is (-ln(3)+13ln(3/4))/2

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integral from 0 to 1 of (x-9)/(x^2-5x+6)

\int_{0}^{1} \frac{x - 9}{x ^{2} - 5 x + 6} d x

\frac{- ln ( 3 ) + 13 ln ( \frac{3}{4} )}{2}

Decimal Notation

- 2.41923 \dots

Solution steps

\int_{0}^{1} \frac{x - 9}{x ^{2} - 5 x + 6} d x

Apply u-substitution

: \int_{- 9}^{- 8} \frac{u}{u ^{2} + 13 u + 42} d u

= \int_{- 9}^{- 8} \frac{u}{u ^{2} + 13 u + 42} d u

Complete the square

u^{2} + 13 u + 42 : (u + \frac{13}{2})^{2} - \frac{1}{4}

= \int_{- 9}^{- 8} \frac{u}{( u + \frac{13}{2} ) ^{2} - \frac{1}{4}} d u

Apply u-substitution

: \int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 ( 2 v - 13 )}{4 v ^{2} - 1} d v

= \int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 ( 2 v - 13 )}{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 - 13}{4 v ^{2} - 1} d v

Expand

\frac{2 v - 13}{4 v ^{2} - 1} : \frac{2 v}{4 v ^{2} - 1} - \frac{13}{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{13}{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{13}{4 v ^{2} - 1} d v = - \frac{13 ln ( \frac{3}{4} )}{4}

= 2 (- \frac{1}{4} ln (3) - (- \frac{13 ln ( \frac{3}{4} )}{4}))

Simplify

2 (- \frac{1}{4} ln (3) - (- \frac{13 ln ( \frac{3}{4} )}{4})) : \frac{- ln ( 3 ) + 13 ln ( \frac{3}{4} )}{2}

= \frac{- ln ( 3 ) + 13 ln ( \frac{3}{4} )}{2}

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

y = 5 x, y = 2 x - 5, y = 0

x \to 4 lim (6 x^{2} - 10)

y^{^{'}} = x y + x - 2 y - 2

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

What is the integral from 0 to 1 of (x-9)/(x^2-5x+6) ?
The integral from 0 to 1 of (x-9)/(x^2-5x+6) is (-ln(3)+13ln(3/4))/2