What is the integral of 1/(x^2+5x+6) ?

The integral of 1/(x^2+5x+6) is ln|x+2|-ln|x+3|+C

integral of 1/(x^2+5x+6)

\int \frac{1}{x ^{2} + 5 x + 6} d x

ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{1}{x ^{2} + 5 x + 6} d x

Take the partial fraction of

\frac{1}{x ^{2} + 5 x + 6} : \frac{1}{x + 2} - \frac{1}{x + 3}

= \int \frac{1}{x + 2} - \frac{1}{x + 3} 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}{x + 2} d x - \int \frac{1}{x + 3} d x

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

\int \frac{1}{x + 3} d x = ln ∣ x + 3 ∣

= ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣

Add a constant to the solution

= ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣ + C

\frac{d}{d x} ((x^{2} + 2) (x^{3} + 1))

\frac{d}{d x} (ln (x + y - 1))

x^{2} (\frac{d y}{d x}) = y - y x

d er i v a t i v e y = (x + 2)^{2} - 3

\int (3 x^{2} + ln (x)) d x

What is the integral of 1/(x^2+5x+6) ?
The integral of 1/(x^2+5x+6) is ln|x+2|-ln|x+3|+C