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

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

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

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

ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{1}{x + 1} - \frac{1}{x + 2} 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 + 1} d x - \int \frac{1}{x + 2} d x

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

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

= ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + C

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