What is the integral from 4 to 5 of x/(x^2+6x+13) ?

The integral from 4 to 5 of x/(x^2+6x+13) is (ln(68)-ln(53)-3(arctan(4)-arctan(7/2)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 5 of x/(x^2+6x+13)

\int_{4}^{5} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( 68 ) - ln ( 53 ) - 3 ( arctan ( 4 ) - arctan ( \frac{7}{2} ) )}{2}

Decimal Notation

0.07462 \dots

Solution steps

\int_{4}^{5} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

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

Apply u-substitution

: \int_{7}^{8} \frac{u - 3}{u ^{2} + 4} d u

= \int_{7}^{8} \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_{7}^{8} \frac{u}{u ^{2} + 4} d u - \int_{7}^{8} \frac{3}{u ^{2} + 4} d u

\int_{7}^{8} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (68) - ln (53))

\int_{7}^{8} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (4) - arctan (\frac{7}{2}))

= \frac{1}{2} (ln (68) - ln (53)) - \frac{3}{2} (arctan (4) - arctan (\frac{7}{2}))

Simplify

\frac{1}{2} (ln (68) - ln (53)) - \frac{3}{2} (arctan (4) - arctan (\frac{7}{2})) : \frac{ln ( 68 ) - ln ( 53 ) - 3 ( arctan ( 4 ) - arctan ( \frac{7}{2} ) )}{2}

= \frac{ln ( 68 ) - ln ( 53 ) - 3 ( arctan ( 4 ) - arctan ( \frac{7}{2} ) )}{2}

\frac{\partial}{\partial x} (6 x)

\frac{\partial}{\partial x} ((A^{2} x^{2} + B^{2} y^{2})^{4})

\frac{\partial}{\partial x} (a (x - b))

\int \frac{2 x + 5}{x} d x

\int_{0}^{16} x 16 - x d x

What is the integral from 4 to 5 of x/(x^2+6x+13) ?
The integral from 4 to 5 of x/(x^2+6x+13) is (ln(68)-ln(53)-3(arctan(4)-arctan(7/2)))/2