What is the integral from 0 to 2 of x/(x^2+4x+8) ?

The integral from 0 to 2 of x/(x^2+4x+8) is 1/2 ln(5)-1/2 ln(2)-arctan(2)+pi/4

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integral from 0 to 2 of x/(x^2+4x+8)

\int_{0}^{2} \frac{x}{x ^{2} + 4 x + 8} d x

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

Decimal Notation

0.13639 \dots

Solution steps

\int_{0}^{2} \frac{x}{x ^{2} + 4 x + 8} d x

Complete the square

x^{2} + 4 x + 8 : (x + 2)^{2} + 4

= \int_{0}^{2} \frac{x}{( x + 2 ) ^{2} + 4} d x

Apply u-substitution

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

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

Apply Integral Substitution

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

Expand

\frac{v - 1}{v ^{2} + 1} : \frac{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

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

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

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

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

Simplify

\frac{1}{2} (ln (5) - ln (2)) - (arctan (2) - \frac{π}{4}) : \frac{1}{2} ln (5) - \frac{1}{2} ln (2) - arctan (2) + \frac{π}{4}

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

\int 4 e^{- t} d t

y = e^{6 e^{x}}

\frac{d}{d x} (- 5 x^{- 2})

x = y^{2}, x = 4 y - y^{2}

\int 2 cos (x) d x

What is the integral from 0 to 2 of x/(x^2+4x+8) ?
The integral from 0 to 2 of x/(x^2+4x+8) is 1/2 ln(5)-1/2 ln(2)-arctan(2)+pi/4