What is the integral of 5/(x^2+7x+12) ?

The integral of 5/(x^2+7x+12) is 5(ln|x+3|-ln|x+4|)+C

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integral of 5/(x^2+7x+12)

\int \frac{5}{x ^{2} + 7 x + 12} d x

5 (ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣) + C

Solution steps

\int \frac{5}{x ^{2} + 7 x + 12} d x

Take the constant out:

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

= 5 \cdot \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}

= 5 \cdot \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

= 5 (\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 ∣

= 5 (ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣)

Add a constant to the solution

= 5 (ln ∣ x + 3 ∣ - ln ∣ x + 4 ∣) + C

\int \frac{3}{x - 4} d x

\int e^{\frac{x}{b}} d x

\int \frac{t ^{2} - 1}{t ^{2} + 1} d t

\int 5 - 0.03 x + 0.0004 x^{2} d x

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

What is the integral of 5/(x^2+7x+12) ?
The integral of 5/(x^2+7x+12) is 5(ln|x+3|-ln|x+4|)+C