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

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

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

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

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

Decimal Notation

0.11778 \dots

Solution steps

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

Complete the square

x^{2} + 6 x + 8 : (x + 3)^{2} - 1

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

Apply u-substitution

: \int_{3}^{5} \frac{u - 3}{u ^{2} - 1} d u

= \int_{3}^{5} \frac{u - 3}{u ^{2} - 1} d u

Expand

\frac{u - 3}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} - \frac{3}{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

= \int_{3}^{5} \frac{u}{u ^{2} - 1} d u - \int_{3}^{5} \frac{3}{u ^{2} - 1} d u

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

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

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

Simplify

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

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

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