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

The integral from 2 to 5 of x/(x^2+6x+13) is (ln(68)-ln(29)-3(arctan(4)-arctan(5/2)))/2

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

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

\frac{ln ( 68 ) - ln ( 29 ) - 3 ( arctan ( 4 ) - arctan ( \frac{5}{2} ) )}{2}

Decimal

0.22281 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{5}^{8} \frac{u - 3}{u ^{2} + 4} d u

= \int_{5}^{8} \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_{5}^{8} \frac{u}{u ^{2} + 4} d u - \int_{5}^{8} \frac{3}{u ^{2} + 4} d u

\int_{5}^{8} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (68) - ln (29))

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

= \frac{1}{2} (ln (68) - ln (29)) - \frac{3}{2} (arctan (4) - arctan (\frac{5}{2}))

Simplify

\frac{1}{2} (ln (68) - ln (29)) - \frac{3}{2} (arctan (4) - arctan (\frac{5}{2})) : \frac{ln ( 68 ) - ln ( 29 ) - 3 ( arctan ( 4 ) - arctan ( \frac{5}{2} ) )}{2}

= \frac{ln ( 68 ) - ln ( 29 ) - 3 ( arctan ( 4 ) - arctan ( \frac{5}{2} ) )}{2}

\frac{\partial}{\partial x} (5 y^{6} cos (5 x))

f (x) = \frac{6 - x ^{2}}{6 + x ^{2}}

\frac{d ^{3}}{d x ^{3}} (ln (1 + x))

k = 1 \sum \infty \frac{k}{e ^{7 k}}

\frac{( e ^{- s} )}{s}

What is the integral from 2 to 5 of x/(x^2+6x+13) ?
The integral from 2 to 5 of x/(x^2+6x+13) is (ln(68)-ln(29)-3(arctan(4)-arctan(5/2)))/2