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

The integral from 5 to 6 of x/(x^2+6x+13) is (ln(5/4)-3(arctan(9/2)-arctan(4)))/2

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

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

\frac{ln ( \frac{5}{4} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 4 ) )}{2}

Decimal

0.07210 \dots

Solution steps

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

Complete the square

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

= \int_{5}^{6} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

= \int_{8}^{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_{8}^{9} \frac{u}{u ^{2} + 4} d u - \int_{8}^{9} \frac{3}{u ^{2} + 4} d u

\int_{8}^{9} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{5}{4})

\int_{8}^{9} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{9}{2}) - arctan (4))

= \frac{1}{2} ln (\frac{5}{4}) - \frac{3}{2} (arctan (\frac{9}{2}) - arctan (4))

Simplify

\frac{1}{2} ln (\frac{5}{4}) - \frac{3}{2} (arctan (\frac{9}{2}) - arctan (4)) : \frac{ln ( \frac{5}{4} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 4 ) )}{2}

= \frac{ln ( \frac{5}{4} ) - 3 ( arctan ( \frac{9}{2} ) - arctan ( 4 ) )}{2}

d er i v a t i v e e^{a x} sin (b x)

\frac{\partial}{\partial y} (e^{2 x} cos (y))

\int e^{- x} cos^{2} (x) d x

\int (x 9 - x^{2}) d x

\frac{d y}{d x} = e^{2 y + 2 x}

What is the integral from 5 to 6 of x/(x^2+6x+13) ?
The integral from 5 to 6 of x/(x^2+6x+13) is (ln(5/4)-3(arctan(9/2)-arctan(4)))/2