What is the integral from 1 to 5 of x/(x^2+6x+13) ?

The integral from 1 to 5 of x/(x^2+6x+13) is (ln(17/5)-3(arctan(4)-arctan(2)))/2

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integral from 1 to 5 of x/(x^2+6x+13)

\int_{1}^{5} \frac{x}{x ^{2} + 6 x + 13} d x

\frac{ln ( \frac{17}{5} ) - 3 ( arctan ( 4 ) - arctan ( 2 ) )}{2}

Decimal Notation

0.28388 \dots

Solution steps

\int_{1}^{5} \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

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

= \int_{1}^{5} \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int_{4}^{8} \frac{u - 3}{u ^{2} + 4} d u

= \int_{4}^{8} \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}^{8} \frac{u}{u ^{2} + 4} d u - \int_{4}^{8} \frac{3}{u ^{2} + 4} d u

\int_{4}^{8} \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln (\frac{17}{5})

\int_{4}^{8} \frac{3}{u ^{2} + 4} d u = \frac{3}{2} (arctan (4) - arctan (2))

= \frac{1}{2} ln (\frac{17}{5}) - \frac{3}{2} (arctan (4) - arctan (2))

Simplify

\frac{1}{2} ln (\frac{17}{5}) - \frac{3}{2} (arctan (4) - arctan (2)) : \frac{ln ( \frac{17}{5} ) - 3 ( arctan ( 4 ) - arctan ( 2 ) )}{2}

= \frac{ln ( \frac{17}{5} ) - 3 ( arctan ( 4 ) - arctan ( 2 ) )}{2}

\int \frac{1}{1 + e ^{12 x}} d x

(7, 0), (- 1, 2)

e^{- (t - 3)^{2}}

\frac{d}{d x} (arctan (sin (x)))

\frac{\partial}{\partial y} (x^{3} y^{4})

What is the integral from 1 to 5 of x/(x^2+6x+13) ?
The integral from 1 to 5 of x/(x^2+6x+13) is (ln(17/5)-3(arctan(4)-arctan(2)))/2