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

The integral from 2 to 7 of x/(x^2+6x+13) is (ln(104)-ln(29)-3(arctan(5)-arctan(5/2)))/2

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

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

\frac{ln ( 104 ) - ln ( 29 ) - 3 ( arctan ( 5 ) - arctan ( \frac{5}{2} ) )}{2}

Decimal

0.36388 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{5}^{10} \frac{u - 3}{u ^{2} + 4} d u

= \int_{5}^{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_{5}^{10} \frac{u}{u ^{2} + 4} d u - \int_{5}^{10} \frac{3}{u ^{2} + 4} d u

\int_{5}^{10} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (104) - ln (29))

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

= \frac{1}{2} (ln (104) - ln (29)) - \frac{3}{2} (arctan (5) - arctan (\frac{5}{2}))

Simplify

\frac{1}{2} (ln (104) - ln (29)) - \frac{3}{2} (arctan (5) - arctan (\frac{5}{2})) : \frac{ln ( 104 ) - ln ( 29 ) - 3 ( arctan ( 5 ) - arctan ( \frac{5}{2} ) )}{2}

= \frac{ln ( 104 ) - ln ( 29 ) - 3 ( arctan ( 5 ) - arctan ( \frac{5}{2} ) )}{2}

\frac{d}{d x} (e^{\frac{x}{4}})

\int \frac{x ^{2} - 25}{2 x ^{2}} d x

\int x^{2} \cdot x^{3} + 3 d x

\frac{d y}{d x} = \frac{( 9 x ^{11} x ^{23} )}{y}

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

What is the integral from 2 to 7 of x/(x^2+6x+13) ?
The integral from 2 to 7 of x/(x^2+6x+13) is (ln(104)-ln(29)-3(arctan(5)-arctan(5/2)))/2