What is the integral from 1 to 8 of x/(x^2+4x+13) ?

The integral from 1 to 8 of x/(x^2+4x+13) is pi/6+(-ln(2)+ln(109/9))/2-2/3 arctan(10/3)

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integral from 1 to 8 of x/(x^2+4x+13)

\int_{1}^{8} \frac{x}{x ^{2} + 4 x + 13} d x

\frac{π}{6} + \frac{- ln ( 2 ) + ln ( \frac{109}{9} )}{2} - \frac{2}{3} arctan (\frac{10}{3})

Decimal

0.57119 \dots

Solution steps

\int_{1}^{8} \frac{x}{x ^{2} + 4 x + 13} d x

Complete the square

x^{2} + 4 x + 13 : (x + 2)^{2} + 9

= \int_{1}^{8} \frac{x}{( x + 2 ) ^{2} + 9} d x

Apply u-substitution

= \int_{3}^{10} \frac{u - 2}{u ^{2} + 9} d u

Apply Integral Substitution

= \int_{1}^{\frac{10}{3}} \frac{3 v - 2}{3 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{1}^{\frac{10}{3}} \frac{3 v - 2}{v ^{2} + 1} d v

Expand

\frac{3 v - 2}{v ^{2} + 1} : \frac{3 v}{v ^{2} + 1} - \frac{2}{v ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int_{1}^{\frac{10}{3}} \frac{3 v}{v ^{2} + 1} d v - \int_{1}^{\frac{10}{3}} \frac{2}{v ^{2} + 1} d v)

\int_{1}^{\frac{10}{3}} \frac{3 v}{v ^{2} + 1} d v = \frac{3}{2} (ln (\frac{109}{9}) - ln (2))

\int_{1}^{\frac{10}{3}} \frac{2}{v ^{2} + 1} d v = 2 (arctan (\frac{10}{3}) - \frac{π}{4})

= \frac{1}{3} (\frac{3}{2} (ln (\frac{109}{9}) - ln (2)) - 2 (arctan (\frac{10}{3}) - \frac{π}{4}))

Simplify

\frac{1}{3} (\frac{3}{2} (ln (\frac{109}{9}) - ln (2)) - 2 (arctan (\frac{10}{3}) - \frac{π}{4})) : \frac{π}{6} + \frac{- ln ( 2 ) + ln ( \frac{109}{9} )}{2} - \frac{2}{3} arctan (\frac{10}{3})

= \frac{π}{6} + \frac{- ln ( 2 ) + ln ( \frac{109}{9} )}{2} - \frac{2}{3} arctan (\frac{10}{3})

What is the integral from 1 to 8 of x/(x^2+4x+13) ?
The integral from 1 to 8 of x/(x^2+4x+13) is pi/6+(-ln(2)+ln(109/9))/2-2/3 arctan(10/3)