What is the integral from 4 to 7 of x/(x^2+6x+13) ?

The integral from 4 to 7 of x/(x^2+6x+13) is (ln(104)-ln(53)-3(arctan(5)-arctan(7/2)))/2

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integral from 4 to 7 of x/(x^2+6x+13)

\int_{4}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( 104 ) - ln ( 53 ) - 3 ( arctan ( 5 ) - arctan ( \frac{7}{2} ) )}{2}

Solution steps

\int_{4}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

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

Apply u-substitution

: \int_{7}^{10} \frac{u - 3}{u ^{2} + 4} d u

= \int_{7}^{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_{7}^{10} \frac{u}{u ^{2} + 4} d u - \int_{7}^{10} \frac{3}{u ^{2} + 4} d u

\int_{7}^{10} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (104) - ln (53))

\int_{7}^{10} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (5) - arctan (\frac{7}{2}))

= \frac{1}{2} (ln (104) - ln (53)) - \frac{3}{2} (arctan (5) - arctan (\frac{7}{2}))

Simplify

\frac{1}{2} (ln (104) - ln (53)) - \frac{3}{2} (arctan (5) - arctan (\frac{7}{2})) : \frac{ln ( 104 ) - ln ( 53 ) - 3 ( arctan ( 5 ) - arctan ( \frac{7}{2} ) )}{2}

= \frac{ln ( 104 ) - ln ( 53 ) - 3 ( arctan ( 5 ) - arctan ( \frac{7}{2} ) )}{2}

f (x) = 2 x^{2} + 2 x, at x = - 3

\int x e - x d x

\int r^{2} sin (θ) d r

\int sin^{2} (\frac{1}{2}) x d x

(- 4, 2), (- 9, - 10)

What is the integral from 4 to 7 of x/(x^2+6x+13) ?
The integral from 4 to 7 of x/(x^2+6x+13) is (ln(104)-ln(53)-3(arctan(5)-arctan(7/2)))/2