What is the integral of (x+3)/(x^2+6x) ?

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

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integral of (x+3)/(x^2+6x)

\int \frac{x + 3}{x ^{2} + 6 x} d x

\frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{x + 3}{x ^{2} + 6 x} d x

Take the partial fraction of

\frac{x + 3}{x ^{2} + 6 x} : \frac{1}{2 x} + \frac{1}{2 ( x + 6 )}

= \int \frac{1}{2 x} + \frac{1}{2 ( x + 6 )} 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} d x + \int \frac{1}{2 ( x + 6 )} d x

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

\int \frac{1}{2 ( x + 6 )} d x = \frac{1}{2} ln ∣ x + 6 ∣

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 6 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 6 ∣ + C

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