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

The integral from 5 to 6 of x/(x^2+6x+13) is (ln(5/4)-3(arctan(9/2)-arctan(4)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{ln ( \frac{5}{4} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 4 ) )}{2}

Decimal Notation

0.07210 \dots

Solution steps

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

Complete the square

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

= \int_{5}^{6} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{8}^{9} \frac{u - 3}{u ^{2} + 4} d u

= \int_{8}^{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_{8}^{9} \frac{u}{u ^{2} + 4} d u - \int_{8}^{9} \frac{3}{u ^{2} + 4} d u

\int_{8}^{9} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{5}{4})

\int_{8}^{9} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{9}{2}) - arctan (4))

= \frac{1}{2} ln (\frac{5}{4}) - \frac{3}{2} (arctan (\frac{9}{2}) - arctan (4))

Simplify

\frac{1}{2} ln (\frac{5}{4}) - \frac{3}{2} (arctan (\frac{9}{2}) - arctan (4)) : \frac{ln ( \frac{5}{4} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 4 ) )}{2}

= \frac{ln ( \frac{5}{4} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 4 ) )}{2}

e^{a x} sin (b x)

x \to 3 lim (e^{\frac{2}{3 - x}})

\frac{\partial}{\partial y} (e^{2 x} cos (y))

\int e^{- x} cos^{2} (x) d x

\int (x 9 - x^{2}) d x

What is the integral from 5 to 6 of x/(x^2+6x+13) ?
The integral from 5 to 6 of x/(x^2+6x+13) is (ln(5/4)-3(arctan(9/2)-arctan(4)))/2