What is the integral from 5 to 7 of x/(x^2+6x+13) ?

The integral from 5 to 7 of x/(x^2+6x+13) is (ln(26/17)-3(arctan(5)-arctan(4)))/2

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integral from 5 to 7 of x/(x^2+6x+13)

\int_{5}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( \frac{26}{17} ) - 3 ( arctan ( 5 ) - arctan ( 4 ) )}{2}

Decimal Notation

0.14106 \dots

Solution steps

\int_{5}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

= \int_{5}^{7} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{8}^{10} \frac{u - 3}{u ^{2} + 4} d u

= \int_{8}^{10} \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}^{10} \frac{u}{u ^{2} + 4} d u - \int_{8}^{10} \frac{3}{u ^{2} + 4} d u

\int_{8}^{10} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{26}{17})

\int_{8}^{10} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (5) - arctan (4))

= \frac{1}{2} ln (\frac{26}{17}) - \frac{3}{2} (arctan (5) - arctan (4))

Simplify

\frac{1}{2} ln (\frac{26}{17}) - \frac{3}{2} (arctan (5) - arctan (4)) : \frac{ln ( \frac{26}{17} ) - 3 ( arctan ( 5 ) - arctan ( 4 ) )}{2}

= \frac{ln ( \frac{26}{17} ) - 3 ( arctan ( 5 ) - arctan ( 4 ) )}{2}

(2 x^{2} + 2)^{3} (x^{2} - 1)^{2}

(2 x^{\frac{3}{2}})^{^{'}}

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 7

\frac{\partial}{\partial x} (x^{2} - y^{3} - x^{2} y + y)

\frac{\partial}{\partial x} (- 2 (1 - x) - 40 x (y - x^{2}))

What is the integral from 5 to 7 of x/(x^2+6x+13) ?
The integral from 5 to 7 of x/(x^2+6x+13) is (ln(26/17)-3(arctan(5)-arctan(4)))/2