What is the integral from 3 to 8 of x/(x^2+6x+13) ?

The integral from 3 to 8 of x/(x^2+6x+13) is (ln(25/8)-3(arctan(11/2)-arctan(3)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 3 to 8 of x/(x^2+6x+13)

\int_{3}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( \frac{25}{8} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 3 ) )}{2}

Decimal Notation

0.35687 \dots

Solution steps

\int_{3}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

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

Apply u-substitution

: \int_{6}^{11} \frac{u - 3}{u ^{2} + 4} d u

= \int_{6}^{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_{6}^{11} \frac{u}{u ^{2} + 4} d u - \int_{6}^{11} \frac{3}{u ^{2} + 4} d u

\int_{6}^{11} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{25}{8})

\int_{6}^{11} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{11}{2}) - arctan (3))

= \frac{1}{2} ln (\frac{25}{8}) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (3))

Simplify

\frac{1}{2} ln (\frac{25}{8}) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (3)) : \frac{ln ( \frac{25}{8} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 3 ) )}{2}

= \frac{ln ( \frac{25}{8} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 3 ) )}{2}

(12 y^{2} t^{2} + 10 y) d y + (8 y^{3} t) d t = 0

f (x) = 11 x

\int \frac{6 x - 5}{3 x ^{2} + 5 x} d x

y = 2 x^{3} - 9 x, y = 3 x^{2} - 10

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x} cos (2 x)

What is the integral from 3 to 8 of x/(x^2+6x+13) ?
The integral from 3 to 8 of x/(x^2+6x+13) is (ln(25/8)-3(arctan(11/2)-arctan(3)))/2