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

The integral from 2 to 8 of x/(x^2+6x+13) is (3ln(5)-ln(29)-3(arctan(11/2)-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 8 of x/(x^2+6x+13)

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

\frac{3 ln ( 5 ) - ln ( 29 ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( \frac{5}{2} ) )}{2}

Decimal Notation

0.42952 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{5}^{11} \frac{u - 3}{u ^{2} + 4} d u

= \int_{5}^{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_{5}^{11} \frac{u}{u ^{2} + 4} d u - \int_{5}^{11} \frac{3}{u ^{2} + 4} d u

\int_{5}^{11} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (3 ln (5) - ln (29))

\int_{5}^{11} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{11}{2}) - arctan (\frac{5}{2}))

= \frac{1}{2} (3 ln (5) - ln (29)) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (\frac{5}{2}))

Simplify

\frac{1}{2} (3 ln (5) - ln (29)) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (\frac{5}{2})) : \frac{3 ln ( 5 ) - ln ( 29 ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( \frac{5}{2} ) )}{2}

= \frac{3 ln ( 5 ) - ln ( 29 ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( \frac{5}{2} ) )}{2}

f (x) = \frac{1}{4 x ^{2}}

\int x^{3} y^{2} d y

\int \frac{1}{x 4 + x ^{2}} d x

y^{^{'}} = k y

y^{^{''}} - y^{^{'}} - 2 y = 4 x^{2}, y (0) = - 1, y^{^{'}} (0) = 1

What is the integral from 2 to 8 of x/(x^2+6x+13) ?
The integral from 2 to 8 of x/(x^2+6x+13) is (3ln(5)-ln(29)-3(arctan(11/2)-arctan(5/2)))/2