What is the integral from 3 to 6 of x/(x^2+4x+20) ?

The integral from 3 to 6 of x/(x^2+4x+20) is 1/2 (ln(5)-ln(41/16)-arctan(2)+arctan(5/4))

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integral from 3 to 6 of x/(x^2+4x+20)

\int_{3}^{6} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{2} (ln (5) - ln (\frac{41}{16}) - arctan (2) + arctan (\frac{5}{4}))

Decimal

0.22868 \dots

Solution steps

\int_{3}^{6} \frac{x}{x ^{2} + 4 x + 20} d x

Complete the square

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int_{3}^{6} \frac{x}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

= \int_{5}^{8} \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

= \int_{\frac{5}{4}}^{2} \frac{2 v - 1}{2 ( 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}{2} \cdot \int_{\frac{5}{4}}^{2} \frac{2 v - 1}{v ^{2} + 1} d v

Expand

\frac{2 v - 1}{v ^{2} + 1} : \frac{2 v}{v ^{2} + 1} - \frac{1}{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}{2} (\int_{\frac{5}{4}}^{2} \frac{2 v}{v ^{2} + 1} d v - \int_{\frac{5}{4}}^{2} \frac{1}{v ^{2} + 1} d v)

\int_{\frac{5}{4}}^{2} \frac{2 v}{v ^{2} + 1} d v = ln (5) - ln (\frac{41}{16})

\int_{\frac{5}{4}}^{2} \frac{1}{v ^{2} + 1} d v = arctan (2) - arctan (\frac{5}{4})

= \frac{1}{2} (ln (5) - ln (\frac{41}{16}) - (arctan (2) - arctan (\frac{5}{4})))

Simplify

= \frac{1}{2} (ln (5) - ln (\frac{41}{16}) - arctan (2) + arctan (\frac{5}{4}))

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\frac{\partial}{\partial x} ((x - 1)^{\frac{2}{3}} (y - 4))

\frac{d}{d y} (y ln (y))

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What is the integral from 3 to 6 of x/(x^2+4x+20) ?
The integral from 3 to 6 of x/(x^2+4x+20) is 1/2 (ln(5)-ln(41/16)-arctan(2)+arctan(5/4))