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

The integral from 3 to 4 of x/(x^2+4x+20) is 1/2 (ln(52/41)-arctan(3/2)+arctan(5/4))

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

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

\frac{1}{2} (ln (\frac{52}{41}) - arctan (\frac{3}{2}) + arctan (\frac{5}{4}))

Decimal Notation

0.07546 \dots

Solution steps

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

Complete the square

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

= \int_{3}^{4} \frac{x}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int_{5}^{6} \frac{u - 2}{u ^{2} + 16} d u

= \int_{5}^{6} \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

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

\int_{\frac{5}{4}}^{\frac{3}{2}} \frac{2 v}{v ^{2} + 1} d v = ln (\frac{52}{41})

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

= \frac{1}{2} (ln (\frac{52}{41}) - (arctan (\frac{3}{2}) - arctan (\frac{5}{4})))

Simplify

= \frac{1}{2} (ln (\frac{52}{41}) - arctan (\frac{3}{2}) + arctan (\frac{5}{4}))

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

t \to 0 lim (\frac{e ^{6 t} - 1}{sin ( t )})

θ \to + 0 lim (\frac{sin ( 8 θ )}{θ})

\int e^{- 9 x} d x

5 ln (x), x ln (x)

What is the integral from 3 to 4 of x/(x^2+4x+20) ?
The integral from 3 to 4 of x/(x^2+4x+20) is 1/2 (ln(52/41)-arctan(3/2)+arctan(5/4))