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

The integral from 4 to 8 of x/(x^2+4x+20) is 1/2 (ln(29/4)-ln(13/4)-arctan(5/2)+arctan(3/2))

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

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

\frac{1}{2} (ln (\frac{29}{4}) - ln (\frac{13}{4}) - arctan (\frac{5}{2}) + arctan (\frac{3}{2}))

Decimal Notation

0.29742 \dots

Solution steps

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

Complete the square

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

= \int_{4}^{8} \frac{x}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int_{6}^{10} \frac{u - 2}{u ^{2} + 16} d u

= \int_{6}^{10} \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

= \int_{\frac{3}{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_{\frac{3}{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_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 v}{v ^{2} + 1} d v - \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{v ^{2} + 1} d v)

\int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 v}{v ^{2} + 1} d v = ln (\frac{29}{4}) - ln (\frac{13}{4})

\int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{v ^{2} + 1} d v = arctan (\frac{5}{2}) - arctan (\frac{3}{2})

= \frac{1}{2} (ln (\frac{29}{4}) - ln (\frac{13}{4}) - (arctan (\frac{5}{2}) - arctan (\frac{3}{2})))

Simplify

= \frac{1}{2} (ln (\frac{29}{4}) - ln (\frac{13}{4}) - arctan (\frac{5}{2}) + arctan (\frac{3}{2}))

\frac{\partial}{\partial x} (x^{x})

x \to \infty lim (9.35 e^{(- 0.4 x)} + 5.52)

x \to - 2 lim (\frac{x ^{2} + 2}{x + 2})

\frac{1}{x} lo g_{10} (1 + x)

\int \frac{7 x - 1}{x ^{2} - x - 6} d x

What is the integral from 4 to 8 of x/(x^2+4x+20) ?
The integral from 4 to 8 of x/(x^2+4x+20) is 1/2 (ln(29/4)-ln(13/4)-arctan(5/2)+arctan(3/2))