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

The integral from 2 to 6 of x/(x^2+6x+13) is (ln(85)-ln(29)-3(arctan(9/2)-arctan(5/2)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{ln ( 85 ) - ln ( 29 ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( \frac{5}{2} ) )}{2}

Decimal

0.29492 \dots

Solution steps

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

Complete the square

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

= \int_{2}^{6} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{5}^{9} \frac{u - 3}{u ^{2} + 4} d u

= \int_{5}^{9} \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}^{9} \frac{u}{u ^{2} + 4} d u - \int_{5}^{9} \frac{3}{u ^{2} + 4} d u

\int_{5}^{9} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (85) - ln (29))

\int_{5}^{9} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{9}{2}) - arctan (\frac{5}{2}))

= \frac{1}{2} (ln (85) - ln (29)) - \frac{3}{2} (arctan (\frac{9}{2}) - arctan (\frac{5}{2}))

Simplify

\frac{1}{2} (ln (85) - ln (29)) - \frac{3}{2} (arctan (\frac{9}{2}) - arctan (\frac{5}{2})) : \frac{ln ( 85 ) - ln ( 29 ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( \frac{5}{2} ) )}{2}

= \frac{ln ( 85 ) - ln ( 29 ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( \frac{5}{2} ) )}{2}

\int - \frac{1}{x ^{2} + 1} d x

y = x^{7}, - 2, 3

t \to 0 lim (\frac{8}{t} - \frac{8}{t ^{2} + t})

tan (y)

y^{^{'}} = \frac{9 x}{y + x ^{2} y}, y (0) = - 5

What is the integral from 2 to 6 of x/(x^2+6x+13) ?
The integral from 2 to 6 of x/(x^2+6x+13) is (ln(85)-ln(29)-3(arctan(9/2)-arctan(5/2)))/2