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

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

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

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

ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣ + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{x ^{2} + 6 x + 8} d x

Take the partial fraction of

\frac{1}{x ^{2} + 6 x + 8} : \frac{1}{2 ( x + 2 )} - \frac{1}{2 ( x + 4 )}

= 2 \cdot \int \frac{1}{2 ( x + 2 )} - \frac{1}{2 ( x + 4 )} d x

Apply the Sum Rule:

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

= 2 (\int \frac{1}{2 ( x + 2 )} d x - \int \frac{1}{2 ( x + 4 )} d x)

\int \frac{1}{2 ( x + 2 )} d x = \frac{1}{2} ln ∣ x + 2 ∣

\int \frac{1}{2 ( x + 4 )} d x = \frac{1}{2} ln ∣ x + 4 ∣

= 2 (\frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 4 ∣)

Simplify

2 (\frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 4 ∣) : ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣

= ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣

Add a constant to the solution

= ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣ + C

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