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

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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

y = cos (e^{- t^{2}})

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

\frac{d}{d t} (0.0225 t e^{- 0.0467 t})

\int \frac{x}{9 - 64 x ^{2}} d x

\int_{0}^{16} 1 + 3 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