What is the integral from 0 to 1 of (x-7)/(x^2-5x+6) ?

The integral from 0 to 1 of (x-7)/(x^2-5x+6) is (-ln(3)+9ln(3/4))/2

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integral from 0 to 1 of (x-7)/(x^2-5x+6)

\int_{0}^{1} \frac{x - 7}{x ^{2} - 5 x + 6} d x

\frac{- ln ( 3 ) + 9 ln ( \frac{3}{4} )}{2}

Decimal Notation

- 1.84387 \dots

Solution steps

\int_{0}^{1} \frac{x - 7}{x ^{2} - 5 x + 6} d x

Apply u-substitution

: \int_{- 7}^{- 6} \frac{u}{u ^{2} + 9 u + 20} d u

= \int_{- 7}^{- 6} \frac{u}{u ^{2} + 9 u + 20} d u

Complete the square

u^{2} + 9 u + 20 : (u + \frac{9}{2})^{2} - \frac{1}{4}

= \int_{- 7}^{- 6} \frac{u}{( u + \frac{9}{2} ) ^{2} - \frac{1}{4}} d u

Apply u-substitution

: \int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 ( 2 v - 9 )}{4 v ^{2} - 1} d v

= \int_{- \frac{5}{2}}^{- \frac{3}{2}} \frac{2 ( 2 v - 9 )}{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 - 9}{4 v ^{2} - 1} d v

Expand

\frac{2 v - 9}{4 v ^{2} - 1} : \frac{2 v}{4 v ^{2} - 1} - \frac{9}{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{9}{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{9}{4 v ^{2} - 1} d v = - \frac{9 ln ( \frac{3}{4} )}{4}

= 2 (- \frac{1}{4} ln (3) - (- \frac{9 ln ( \frac{3}{4} )}{4}))

Simplify

2 (- \frac{1}{4} ln (3) - (- \frac{9 ln ( \frac{3}{4} )}{4})) : \frac{- ln ( 3 ) + 9 ln ( \frac{3}{4} )}{2}

= \frac{- ln ( 3 ) + 9 ln ( \frac{3}{4} )}{2}

y^{^{'}} + 6 y = e^{4 t}

\frac{d}{d x} (sec^{5} (x^{2} - 1))

3 x^{2} + 1

f (x) = x^{\frac{1}{9}}

f (x) = 2 x (2 - 4 x)^{4}, at x = 1

What is the integral from 0 to 1 of (x-7)/(x^2-5x+6) ?
The integral from 0 to 1 of (x-7)/(x^2-5x+6) is (-ln(3)+9ln(3/4))/2