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

The integral from 6 to 8 of x/(x^2+6x+13) is (ln(25/17)-3(arctan(11/2)-arctan(9/2)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{ln ( \frac{25}{17} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( \frac{9}{2} ) )}{2}

Decimal Notation

0.13460 \dots

Solution steps

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

Complete the square

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

= \int_{6}^{8} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{9}^{11} \frac{u - 3}{u ^{2} + 4} d u

= \int_{9}^{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_{9}^{11} \frac{u}{u ^{2} + 4} d u - \int_{9}^{11} \frac{3}{u ^{2} + 4} d u

\int_{9}^{11} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{25}{17})

\int_{9}^{11} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{11}{2}) - arctan (\frac{9}{2}))

= \frac{1}{2} ln (\frac{25}{17}) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (\frac{9}{2}))

Simplify

\frac{1}{2} ln (\frac{25}{17}) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (\frac{9}{2})) : \frac{ln ( \frac{25}{17} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( \frac{9}{2} ) )}{2}

= \frac{ln ( \frac{25}{17} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( \frac{9}{2} ) )}{2}

\frac{\partial}{\partial x} (ln (8 x))

\frac{d}{d x} (\frac{( e ^{x} - 1 )}{( e ^{x} + 1 )})

\frac{\partial}{\partial x} (ln (5 x + 2 y))

x \to 6 lim (1)

ln (1 + x), at (x = 2)

What is the integral from 6 to 8 of x/(x^2+6x+13) ?
The integral from 6 to 8 of x/(x^2+6x+13) is (ln(25/17)-3(arctan(11/2)-arctan(9/2)))/2