What is the integral from 4 to 5 of x/(x^2+4x+20) ?

The integral from 4 to 5 of x/(x^2+4x+20) is 1/2 (ln(5/4)-arctan(7/4)+arctan(3/2))

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integral from 4 to 5 of x/(x^2+4x+20)

\int_{4}^{5} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{2} (ln (\frac{5}{4}) - arctan (\frac{7}{4}) + arctan (\frac{3}{2}))

Decimal Notation

0.07714 \dots

Solution steps

\int_{4}^{5} \frac{x}{x ^{2} + 4 x + 20} d x

Complete the square

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

= \int_{4}^{5} \frac{x}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int_{6}^{7} \frac{u - 2}{u ^{2} + 16} d u

= \int_{6}^{7} \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

= \int_{\frac{3}{2}}^{\frac{7}{4}} \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{7}{4}} \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{7}{4}} \frac{2 v}{v ^{2} + 1} d v - \int_{\frac{3}{2}}^{\frac{7}{4}} \frac{1}{v ^{2} + 1} d v)

\int_{\frac{3}{2}}^{\frac{7}{4}} \frac{2 v}{v ^{2} + 1} d v = ln (\frac{5}{4})

\int_{\frac{3}{2}}^{\frac{7}{4}} \frac{1}{v ^{2} + 1} d v = arctan (\frac{7}{4}) - arctan (\frac{3}{2})

= \frac{1}{2} (ln (\frac{5}{4}) - (arctan (\frac{7}{4}) - arctan (\frac{3}{2})))

Simplify

= \frac{1}{2} (ln (\frac{5}{4}) - arctan (\frac{7}{4}) + arctan (\frac{3}{2}))

\int (- 5 csc^{2} (x)) d x

x cos (2 x), 1

x \to + 1 lim (e^{4 x^{3} - 4 x})

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

\int 30 sin^{3} (x) cos^{2} (x) d x

What is the integral from 4 to 5 of x/(x^2+4x+20) ?
The integral from 4 to 5 of x/(x^2+4x+20) is 1/2 (ln(5/4)-arctan(7/4)+arctan(3/2))