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

The integral from 1 to 2 of x/(x^2+4x+20) is 1/2 (ln(2)-ln(25/16)-pi/4+arctan(3/4))

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

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

\frac{1}{2} (ln (2) - ln (\frac{25}{16}) - \frac{π}{4} + arctan (\frac{3}{4}))

Decimal Notation

0.05248 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{3}^{4} \frac{u - 2}{u ^{2} + 16} d u

= \int_{3}^{4} \frac{u - 2}{u ^{2} + 16} d u

Apply Integral Substitution

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

\int_{\frac{3}{4}}^{1} \frac{2 v}{v ^{2} + 1} d v = ln (2) - ln (\frac{25}{16})

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

= \frac{1}{2} (ln (2) - ln (\frac{25}{16}) - (\frac{π}{4} - arctan (\frac{3}{4})))

Simplify

= \frac{1}{2} (ln (2) - ln (\frac{25}{16}) - \frac{π}{4} + arctan (\frac{3}{4}))

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

\int (sin (y) + y sin (x) + \frac{1}{x}) d x

\frac{d}{d x} (x^{2} \cdot x)

\frac{s}{( s - 3 ) ( s - 6 ) ( s - 2 )}

\int_{0}^{0.4} \frac{( x ^{2} )}{4 - 25 x ^{2}} d x

What is the integral from 1 to 2 of x/(x^2+4x+20) ?
The integral from 1 to 2 of x/(x^2+4x+20) is 1/2 (ln(2)-ln(25/16)-pi/4+arctan(3/4))