What is the integral of (2x+6)/(x^2+4x+8) ?

The integral of (2x+6)/(x^2+4x+8) is ln|x^2+4x+8|+arctan(1/2 (x+2))+C

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integral of (2x+6)/(x^2+4x+8)

\int \frac{2 x + 6}{x ^{2} + 4 x + 8} d x

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

Solution steps

\int \frac{2 x + 6}{x ^{2} + 4 x + 8} d x

Complete the square

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

= \int \frac{2 x + 6}{( x + 2 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{2 u + 2}{u ^{2} + 4} d u

= \int \frac{2 u + 2}{u ^{2} + 4} d u

Expand

\frac{2 u + 2}{u ^{2} + 4} : \frac{2 u}{u ^{2} + 4} + \frac{2}{u ^{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{2 u}{u ^{2} + 4} d u + \int \frac{2}{u ^{2} + 4} d u

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

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

= ln u^{2} + 4 + arctan (\frac{u}{2})

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (2x+6)/(x^2+4x+8) ?
The integral of (2x+6)/(x^2+4x+8) is ln|x^2+4x+8|+arctan(1/2 (x+2))+C