What is the integral of 1/((x+2)(x+3)(x+4)) ?

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

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integral of 1/((x+2)(x+3)(x+4))

\int \frac{1}{( x + 2 ) ( x + 3 ) ( x + 4 )} d x

\frac{1}{2} ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣ + \frac{1}{2} ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{1}{( x + 2 ) ( x + 3 ) ( x + 4 )} d x

Take the partial fraction of

\frac{1}{( x + 2 ) ( x + 3 ) ( x + 4 )} : \frac{1}{2 ( x + 2 )} - \frac{1}{x + 3} + \frac{1}{2 ( x + 4 )}

= \int \frac{1}{2 ( x + 2 )} - \frac{1}{x + 3} + \frac{1}{2 ( 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}{2 ( x + 2 )} d x - \int \frac{1}{x + 3} d x + \int \frac{1}{2 ( x + 4 )} d x

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

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

\int \frac{1}{2 ( x + 4 )} d x = \frac{1}{2} ln ∣ x + 4 ∣

= \frac{1}{2} ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣ + \frac{1}{2} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣ + \frac{1}{2} ln ∣ x + 4 ∣ + C

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