What is the integral of (x+11)/(x^2+4x+8) ?

The integral of (x+11)/(x^2+4x+8) is 1/2 ln|x^2+4x+8|+9/2 arctan((x+2)/2)+C

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integral of (x+11)/(x^2+4x+8)

\int \frac{x + 11}{x ^{2} + 4 x + 8} d x

\frac{1}{2} ln x^{2} + 4 x + 8 + \frac{9}{2} arctan (\frac{x + 2}{2}) + C

Solution steps

\int \frac{x + 11}{x ^{2} + 4 x + 8} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 18 u + 85} d u

= \int \frac{u}{u ^{2} - 18 u + 85} d u

Complete the square

u^{2} - 18 u + 85 : (u - 9)^{2} + 4

= \int \frac{u}{( u - 9 ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{v + 9}{v ^{2} + 4} d v

= \int \frac{v + 9}{v ^{2} + 4} d v

Expand

\frac{v + 9}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{9}{v ^{2} + 4}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{v}{v ^{2} + 4} d v + \int \frac{9}{v ^{2} + 4} d v

\int \frac{v}{v ^{2} + 4} d v = \frac{1}{2} ln v^{2} + 4

\int \frac{9}{v ^{2} + 4} d v = \frac{9}{2} arctan (\frac{v}{2})

= \frac{1}{2} ln v^{2} + 4 + \frac{9}{2} arctan (\frac{v}{2})

Substitute back

= \frac{1}{2} ln (x + 11 - 9)^{2} + 4 + \frac{9}{2} arctan (\frac{x + 11 - 9}{2})

Simplify

\frac{1}{2} ln (x + 11 - 9)^{2} + 4 + \frac{9}{2} arctan (\frac{x + 11 - 9}{2}) : \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{9}{2} arctan (\frac{x + 2}{2})

= \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{9}{2} arctan (\frac{x + 2}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{9}{2} arctan (\frac{x + 2}{2}) + C

n = 1 \sum \infty \frac{1}{4 n + 2}

7 y ln (x) - x y^{^{'}} = 0

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

g (u) = 3 u + 5 u

f (x) = 15 e^{x} + 11 x, at x = 0

What is the integral of (x+11)/(x^2+4x+8) ?
The integral of (x+11)/(x^2+4x+8) is 1/2 ln|x^2+4x+8|+9/2 arctan((x+2)/2)+C