What is the integral from 6 to 8 of x/(x^2+4x+20) ?

The integral from 6 to 8 of x/(x^2+4x+20) is 1/2 (ln(29/4)-ln(5)-arctan(5/2)+arctan(2))

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integral from 6 to 8 of x/(x^2+4x+20)

\int_{6}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{2} (ln (\frac{29}{4}) - ln (5) - arctan (\frac{5}{2}) + arctan (2))

Decimal

0.14421 \dots

Solution steps

\int_{6}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

Complete the square

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

= \int_{6}^{8} \frac{x}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

= \int_{8}^{10} \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

= \int_{2}^{\frac{5}{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_{2}^{\frac{5}{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_{2}^{\frac{5}{2}} \frac{2 v}{v ^{2} + 1} d v - \int_{2}^{\frac{5}{2}} \frac{1}{v ^{2} + 1} d v)

\int_{2}^{\frac{5}{2}} \frac{2 v}{v ^{2} + 1} d v = ln (\frac{29}{4}) - ln (5)

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

= \frac{1}{2} (ln (\frac{29}{4}) - ln (5) - (arctan (\frac{5}{2}) - arctan (2)))

Simplify

= \frac{1}{2} (ln (\frac{29}{4}) - ln (5) - arctan (\frac{5}{2}) + arctan (2))

\frac{\partial}{\partial x} (3 x^{2} - 27)

x \to 0 + lim (x (ln (x))^{2})

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

x \to - 1 lim (x - x^{3})

\frac{d}{d x} ((x y^{3} + x^{2} y)^{3})

What is the integral from 6 to 8 of x/(x^2+4x+20) ?
The integral from 6 to 8 of x/(x^2+4x+20) is 1/2 (ln(29/4)-ln(5)-arctan(5/2)+arctan(2))