What is the integral of 1/(x^2+7x+12) ?

The integral of 1/(x^2+7x+12) is ln|x+3|-ln|x+4|+C

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integral of 1/(x^2+7x+12)

\int \frac{1}{x ^{2} + 7 x + 12} d x

ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{1}{x ^{2} + 7 x + 12} d x

Take the partial fraction of

\frac{1}{x ^{2} + 7 x + 12} : \frac{1}{x + 3} - \frac{1}{x + 4}

= \int \frac{1}{x + 3} - \frac{1}{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}{x + 3} d x - \int \frac{1}{x + 4} d x

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

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

= ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣

Add a constant to the solution

= ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣ + C

What is the integral of 1/(x^2+7x+12) ?
The integral of 1/(x^2+7x+12) is ln|x+3|-ln|x+4|+C