What is the integral from 0 to 1 of x/(x^2+4x+13) ?

The integral from 0 to 1 of x/(x^2+4x+13) is (ln(2)-ln(13/9))/2-pi/6+2/3 arctan(2/3)

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integral from 0 to 1 of x/(x^2+4x+13)

\int_{0}^{1} \frac{x}{x ^{2} + 4 x + 13} d x

\frac{ln ( 2 ) - ln ( \frac{13}{9} )}{2} - \frac{π}{6} + \frac{2}{3} arctan (\frac{2}{3})

Decimal Notation

0.03111 \dots

Solution steps

\int_{0}^{1} \frac{x}{x ^{2} + 4 x + 13} d x

Complete the square

x^{2} + 4 x + 13 : (x + 2)^{2} + 9

= \int_{0}^{1} \frac{x}{( x + 2 ) ^{2} + 9} d x

Apply u-substitution

: \int_{2}^{3} \frac{u - 2}{u ^{2} + 9} d u

= \int_{2}^{3} \frac{u - 2}{u ^{2} + 9} d u

Apply Integral Substitution

= \int_{\frac{2}{3}}^{1} \frac{3 v - 2}{3 ( 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}{3} \cdot \int_{\frac{2}{3}}^{1} \frac{3 v - 2}{v ^{2} + 1} d v

Expand

\frac{3 v - 2}{v ^{2} + 1} : \frac{3 v}{v ^{2} + 1} - \frac{2}{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}{3} (\int_{\frac{2}{3}}^{1} \frac{3 v}{v ^{2} + 1} d v - \int_{\frac{2}{3}}^{1} \frac{2}{v ^{2} + 1} d v)

\int_{\frac{2}{3}}^{1} \frac{3 v}{v ^{2} + 1} d v = \frac{3}{2} (ln (2) - ln (\frac{13}{9}))

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

= \frac{1}{3} (\frac{3}{2} (ln (2) - ln (\frac{13}{9})) - 2 (\frac{π}{4} - arctan (\frac{2}{3})))

Simplify

\frac{1}{3} (\frac{3}{2} (ln (2) - ln (\frac{13}{9})) - 2 (\frac{π}{4} - arctan (\frac{2}{3}))) : \frac{ln ( 2 ) - ln ( \frac{13}{9} )}{2} - \frac{π}{6} + \frac{2}{3} arctan (\frac{2}{3})

= \frac{ln ( 2 ) - ln ( \frac{13}{9} )}{2} - \frac{π}{6} + \frac{2}{3} arctan (\frac{2}{3})

y = 2 x, (1, 2)

y = \frac{x ^{4}}{16}

\int x (3 x - 5)^{10} d x

\int \frac{1}{( x ) ln ( x )} d x

\int \frac{8}{1 + 9 r ^{2}} d r

What is the integral from 0 to 1 of x/(x^2+4x+13) ?
The integral from 0 to 1 of x/(x^2+4x+13) is (ln(2)-ln(13/9))/2-pi/6+2/3 arctan(2/3)