What is the integral of (x+5)/(x^2+4x+8) ?

The integral of (x+5)/(x^2+4x+8) is 1/2 ln|x^2+4x+8|+3/2 arctan((x+2)/2)+C

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integral of (x+5)/(x^2+4x+8)

\int \frac{x + 5}{x ^{2} + 4 x + 8} d x

\frac{1}{2} ln x^{2} + 4 x + 8 + \frac{3}{2} arctan (\frac{x + 2}{2}) + C

Solution steps

\int \frac{x + 5}{x ^{2} + 4 x + 8} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 6 u + 13} d u

= \int \frac{u}{u ^{2} - 6 u + 13} d u

Complete the square

u^{2} - 6 u + 13 : (u - 3)^{2} + 4

= \int \frac{u}{( u - 3 ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{v + 3}{v ^{2} + 4} d v

= \int \frac{v + 3}{v ^{2} + 4} d v

Expand

\frac{v + 3}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{3}{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{3}{v ^{2} + 4} d v

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

\int \frac{3}{v ^{2} + 4} d v = \frac{3}{2} arctan (\frac{v}{2})

= \frac{1}{2} ln v^{2} + 4 + \frac{3}{2} arctan (\frac{v}{2})

Substitute back

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

Simplify

\frac{1}{2} ln (x + 5 - 3)^{2} + 4 + \frac{3}{2} arctan (\frac{x + 5 - 3}{2}) : \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{3}{2} arctan (\frac{x + 2}{2})

= \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{3}{2} arctan (\frac{x + 2}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 8 + \frac{3}{2} arctan (\frac{x + 2}{2}) + C

f (x) = e^{3 x}

\frac{t d y}{d t} = y + 5 t^{2} sin (t)

\frac{d}{d x} ((x - 2)^{3} (x^{2} + 9)^{4})

\frac{d}{d x} (\frac{3 a ^{2}}{x ^{2}} - \frac{a}{3 x ^{4}})

f (x) = 24 x

What is the integral of (x+5)/(x^2+4x+8) ?
The integral of (x+5)/(x^2+4x+8) is 1/2 ln|x^2+4x+8|+3/2 arctan((x+2)/2)+C