What is the integral from 0 to 3 of x/(x^2+6x+13) ?

The integral from 0 to 3 of x/(x^2+6x+13) is (ln(40)-ln(13)-3(arctan(3)-arctan(3/2)))/2

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integral from 0 to 3 of x/(x^2+6x+13)

\int_{0}^{3} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( 40 ) - ln ( 13 ) - 3 ( arctan ( 3 ) - arctan ( \frac{3}{2} ) )}{2}

Decimal Notation

0.16258 \dots

Solution steps

\int_{0}^{3} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

= \int_{0}^{3} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

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

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

\int_{3}^{6} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} (ln (40) - ln (13))

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

= \frac{1}{2} (ln (40) - ln (13)) - \frac{3}{2} (arctan (3) - arctan (\frac{3}{2}))

Simplify

\frac{1}{2} (ln (40) - ln (13)) - \frac{3}{2} (arctan (3) - arctan (\frac{3}{2})) : \frac{ln ( 40 ) - ln ( 13 ) - 3 ( arctan ( 3 ) - arctan ( \frac{3}{2} ) )}{2}

= \frac{ln ( 40 ) - ln ( 13 ) - 3 ( arctan ( 3 ) - arctan ( \frac{3}{2} ) )}{2}

3 y^{^{''}} - 5 y^{^{'}} - 2 y = t^{3} - 2 t

f (x) = sin^{2} (3 x)

\frac{d ^{2}}{d x ^{2}} (cos^{2} (5 π x))

\int 2 cos^{2} (x) tan^{3} (x) d x

\frac{d}{d x} (10 100 x - x^{2})

What is the integral from 0 to 3 of x/(x^2+6x+13) ?
The integral from 0 to 3 of x/(x^2+6x+13) is (ln(40)-ln(13)-3(arctan(3)-arctan(3/2)))/2