What is the integral of 1/(x(x+3)(x+1)) ?

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

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integral of 1/(x(x+3)(x+1))

\int \frac{1}{x ( x + 3 ) ( x + 1 )} d x

\frac{1}{3} ln ∣ x ∣ + \frac{1}{6} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{1}{x ( x + 3 ) ( x + 1 )} d x

Take the partial fraction of

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

= \int \frac{1}{3 x} + \frac{1}{6 ( x + 3 )} - \frac{1}{2 ( x + 1 )} 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}{3 x} d x + \int \frac{1}{6 ( x + 3 )} d x - \int \frac{1}{2 ( x + 1 )} d x

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

\int \frac{1}{6 ( x + 3 )} d x = \frac{1}{6} ln ∣ x + 3 ∣

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

= \frac{1}{3} ln ∣ x ∣ + \frac{1}{6} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ x ∣ + \frac{1}{6} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + C

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