What is the integral from 1 to 4 of x/(x^2+6x+13) ?

The integral from 1 to 4 of x/(x^2+6x+13) is (ln(53)-ln(20)-3(arctan(7/2)-arctan(2)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 4 of x/(x^2+6x+13)

\int_{1}^{4} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( 53 ) - ln ( 20 ) - 3 ( arctan ( \frac{7}{2} ) - arctan ( 2 ) )}{2}

Decimal Notation

0.20925 \dots

Solution steps

\int_{1}^{4} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

= \int_{1}^{4} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{4}^{7} \frac{u - 3}{u ^{2} + 4} d u

= \int_{4}^{7} \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_{4}^{7} \frac{u}{u ^{2} + 4} d u - \int_{4}^{7} \frac{3}{u ^{2} + 4} d u

\int_{4}^{7} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (53) - ln (20))

\int_{4}^{7} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{7}{2}) - arctan (2))

= \frac{1}{2} (ln (53) - ln (20)) - \frac{3}{2} (arctan (\frac{7}{2}) - arctan (2))

Simplify

\frac{1}{2} (ln (53) - ln (20)) - \frac{3}{2} (arctan (\frac{7}{2}) - arctan (2)) : \frac{ln ( 53 ) - ln ( 20 ) - 3 ( arctan ( \frac{7}{2} ) - arctan ( 2 ) )}{2}

= \frac{ln ( 53 ) - ln ( 20 ) - 3 ( arctan ( \frac{7}{2} ) - arctan ( 2 ) )}{2}

\frac{\partial}{\partial x} (arctan (3 x))

x \to \infty lim (\frac{4 x ^{6} - x}{x ^{3} + 2})

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

g (x) = x (x - 4)^{3}

x \to 0 lim (\frac{7 x ^{4} - x ^{3}}{x})

What is the integral from 1 to 4 of x/(x^2+6x+13) ?
The integral from 1 to 4 of x/(x^2+6x+13) is (ln(53)-ln(20)-3(arctan(7/2)-arctan(2)))/2