What is the integral from 1 to 3 of x/(x^2+6x+13) ?

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

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integral from 1 to 3 of x/(x^2+6x+13)

\int_{1}^{3} \frac{x}{x ^{2} + 6 x + 13} d x

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

Decimal

0.13372 \dots

Solution steps

\int_{1}^{3} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

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

Apply u-substitution

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

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

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

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

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

Simplify

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

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

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