What is the integral from 3 to 7 of x/(x^2+6x+13) ?

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

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integral from 3 to 7 of x/(x^2+6x+13)

\int_{3}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( \frac{13}{5} ) - 3 ( arctan ( 5 ) - arctan ( 3 ) )}{2}

Decimal Notation

0.29122 \dots

Solution steps

\int_{3}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

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

Apply u-substitution

: \int_{6}^{10} \frac{u - 3}{u ^{2} + 4} d u

= \int_{6}^{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_{6}^{10} \frac{u}{u ^{2} + 4} d u - \int_{6}^{10} \frac{3}{u ^{2} + 4} d u

\int_{6}^{10} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{13}{5})

\int_{6}^{10} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (5) - arctan (3))

= \frac{1}{2} ln (\frac{13}{5}) - \frac{3}{2} (arctan (5) - arctan (3))

Simplify

\frac{1}{2} ln (\frac{13}{5}) - \frac{3}{2} (arctan (5) - arctan (3)) : \frac{ln ( \frac{13}{5} ) - 3 ( arctan ( 5 ) - arctan ( 3 ) )}{2}

= \frac{ln ( \frac{13}{5} ) - 3 ( arctan ( 5 ) - arctan ( 3 ) )}{2}

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

(1 + x^{2}) \frac{d y}{d x} - 2 x y = x^{2} + x^{4}

\frac{\partial}{\partial x} (20 e^{5 x} y)

f (x) = ln (x^{2} - 1)

\frac{d}{d x} (\frac{x ^{2} - 6 x}{( x - 3 ) ^{2}})

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