What is the integral of 1/(x^2+6x+8) ?

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

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integral of 1/(x^2+6x+8)

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

\frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x + 4 ∣ + C

Solution steps

\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 )}

= \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

= \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 ∣

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

Add a constant to the solution

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

\int 2 x - x^{2} d x

\frac{\partial}{\partial x} (sin (π (6 x - 2 y)))

\int_{1}^{\infty} \frac{1}{x ^{2} + 2} d x

x \to - 6 lim (x^{2} - 7)

\int \frac{x - 240}{200} d x

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