What is the integral from 1 to 8 of x/(x^2+4x+20) ?

The integral from 1 to 8 of x/(x^2+4x+20) is 1/2 (ln(116/25)-arctan(5/2)+arctan(3/4))

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integral from 1 to 8 of x/(x^2+4x+20)

\int_{1}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{2} (ln (\frac{116}{25}) - arctan (\frac{5}{2}) + arctan (\frac{3}{4}))

Decimal Notation

0.49396 \dots

Solution steps

\int_{1}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

Complete the square

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

= \int_{1}^{8} \frac{x}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int_{3}^{10} \frac{u - 2}{u ^{2} + 16} d u

= \int_{3}^{10} \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

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

\int_{\frac{3}{4}}^{\frac{5}{2}} \frac{2 v}{v ^{2} + 1} d v = ln (\frac{116}{25})

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

= \frac{1}{2} (ln (\frac{116}{25}) - (arctan (\frac{5}{2}) - arctan (\frac{3}{4})))

Simplify

= \frac{1}{2} (ln (\frac{116}{25}) - arctan (\frac{5}{2}) + arctan (\frac{3}{4}))

\frac{1}{x ^{4}}

\int_{0}^{4} (4 e^{- 0.25 x} - 6 x) d x

\frac{x}{lo g _{e} ( x )}

5 - 2 x

\int e^{x^{2}} \cdot x d x

What is the integral from 1 to 8 of x/(x^2+4x+20) ?
The integral from 1 to 8 of x/(x^2+4x+20) is 1/2 (ln(116/25)-arctan(5/2)+arctan(3/4))