What is the integral from 5 to 8 of x/(x^2+6x+13) ?

The integral from 5 to 8 of x/(x^2+6x+13) is (3ln(5)-ln(68)-3(arctan(11/2)-arctan(4)))/2

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integral from 5 to 8 of x/(x^2+6x+13)

\int_{5}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{3 ln ( 5 ) - ln ( 68 ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 4 ) )}{2}

Decimal Notation

0.20671 \dots

Solution steps

\int_{5}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= \int_{5}^{8} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{8}^{11} \frac{u - 3}{u ^{2} + 4} d u

= \int_{8}^{11} \frac{u - 3}{u ^{2} + 4} d u

Expand

\frac{u - 3}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{3}{u ^{2} + 4}

Apply the Sum Rule:

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

= \int_{8}^{11} \frac{u}{u ^{2} + 4} d u - \int_{8}^{11} \frac{3}{u ^{2} + 4} d u

\int_{8}^{11} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (3 ln (5) - ln (68))

\int_{8}^{11} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{11}{2}) - arctan (4))

= \frac{1}{2} (3 ln (5) - ln (68)) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (4))

Simplify

\frac{1}{2} (3 ln (5) - ln (68)) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (4)) : \frac{3 ln ( 5 ) - ln ( 68 ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 4 ) )}{2}

= \frac{3 ln ( 5 ) - ln ( 68 ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 4 ) )}{2}

\int_{\frac{1}{7}}^{5} 14 x ln (7 x) d x

f (x) = (x + 1) (x^{3} + x - 2)

\int_{1}^{4} \frac{x ^{2} + 6}{5 x - x ^{2}} d x

\int cos^{2} (π) x d x

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

What is the integral from 5 to 8 of x/(x^2+6x+13) ?
The integral from 5 to 8 of x/(x^2+6x+13) is (3ln(5)-ln(68)-3(arctan(11/2)-arctan(4)))/2