What is the integral from 1 to 8 of x/(x^2+6x+13) ?

The integral from 1 to 8 of x/(x^2+6x+13) is (ln(25/4)-3(arctan(11/2)-arctan(2)))/2

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integral from 1 to 8 of x/(x^2+6x+13)

\int_{1}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( \frac{25}{4} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 2 ) )}{2}

Decimal

0.49059 \dots

Solution steps

\int_{1}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

= \int_{1}^{8} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{4}^{11} \frac{u - 3}{u ^{2} + 4} d u

= \int_{4}^{11} \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_{4}^{11} \frac{u}{u ^{2} + 4} d u - \int_{4}^{11} \frac{3}{u ^{2} + 4} d u

\int_{4}^{11} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{25}{4})

\int_{4}^{11} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (\frac{11}{2}) - arctan (2))

= \frac{1}{2} ln (\frac{25}{4}) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (2))

Simplify

\frac{1}{2} ln (\frac{25}{4}) - \frac{3}{2} (arctan (\frac{11}{2}) - arctan (2)) : \frac{ln ( \frac{25}{4} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 2 ) )}{2}

= \frac{ln ( \frac{25}{4} ) - 3 ( arctan ( \frac{11}{2} ) - arctan ( 2 ) )}{2}

x \to 0 lim (ln^{4} (x) x)

\frac{1}{1 - x}, 2

\frac{d}{d x} (x^{2} e^{k x})

\frac{\partial}{\partial x} (ln (4) e^{x})

4 x^{2} \cdot y^{^{'}} = y^{^{'}} + 9 x \cdot e^{- y}

What is the integral from 1 to 8 of x/(x^2+6x+13) ?
The integral from 1 to 8 of x/(x^2+6x+13) is (ln(25/4)-3(arctan(11/2)-arctan(2)))/2