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

The integral from 0 to 1 of x/(x^2+5x+6) is (ln(2)+5ln(8/9))/2

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

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

\frac{ln ( 2 ) + 5 ln ( \frac{8}{9} )}{2}

Decimal

0.05211 \dots

Solution steps

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

Complete the square

x^{2} + 5 x + 6 : (x + \frac{5}{2})^{2} - \frac{1}{4}

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

Apply u-substitution

: \int_{\frac{5}{2}}^{\frac{7}{2}} \frac{2 ( 2 u - 5 )}{4 u ^{2} - 1} d u

= \int_{\frac{5}{2}}^{\frac{7}{2}} \frac{2 ( 2 u - 5 )}{4 u ^{2} - 1} d u

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{7}{2}} \frac{2 u - 5}{4 u ^{2} - 1} d u

Expand

\frac{2 u - 5}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} - \frac{5}{4 u ^{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{7}{2}} \frac{2 u}{4 u ^{2} - 1} d u - \int_{\frac{5}{2}}^{\frac{7}{2}} \frac{5}{4 u ^{2} - 1} d u)

\int_{\frac{5}{2}}^{\frac{7}{2}} \frac{2 u}{4 u ^{2} - 1} d u = \frac{1}{4} ln (2)

\int_{\frac{5}{2}}^{\frac{7}{2}} \frac{5}{4 u ^{2} - 1} d u = - \frac{5 ln ( \frac{8}{9} )}{4}

= 2 (\frac{1}{4} ln (2) - (- \frac{5 ln ( \frac{8}{9} )}{4}))

Simplify

2 (\frac{1}{4} ln (2) - (- \frac{5 ln ( \frac{8}{9} )}{4})) : \frac{ln ( 2 ) + 5 ln ( \frac{8}{9} )}{2}

= \frac{ln ( 2 ) + 5 ln ( \frac{8}{9} )}{2}

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What is the integral from 0 to 1 of x/(x^2+5x+6) ?
The integral from 0 to 1 of x/(x^2+5x+6) is (ln(2)+5ln(8/9))/2