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

The integral from 3 to 6 of x/(x^2+6x+13) is (ln(17/8)-3(arctan(9/2)-arctan(3)))/2

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

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

\frac{ln ( \frac{17}{8} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 3 ) )}{2}

Decimal Notation

0.22226 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{6}^{9} \frac{u - 3}{u ^{2} + 4} d u

= \int_{6}^{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_{6}^{9} \frac{u}{u ^{2} + 4} d u - \int_{6}^{9} \frac{3}{u ^{2} + 4} d u

\int_{6}^{9} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{17}{8})

\int_{6}^{9} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{9}{2}) - arctan (3))

= \frac{1}{2} ln (\frac{17}{8}) - \frac{3}{2} (arctan (\frac{9}{2}) - arctan (3))

Simplify

\frac{1}{2} ln (\frac{17}{8}) - \frac{3}{2} (arctan (\frac{9}{2}) - arctan (3)) : \frac{ln ( \frac{17}{8} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 3 ) )}{2}

= \frac{ln ( \frac{17}{8} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 3 ) )}{2}

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

\frac{d}{d x} (9 tan (x) - 9 cot (x))

\frac{d}{d x} (x (17 - 2 x) (33 - 2 x))

x \to 0 lim (2 x^{2} + 4 x - 2)

1 2^{^{'}}

What is the integral from 3 to 6 of x/(x^2+6x+13) ?
The integral from 3 to 6 of x/(x^2+6x+13) is (ln(17/8)-3(arctan(9/2)-arctan(3)))/2