What is the integral of-1/((z+2)*(z+1)z) ?

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

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integral of-1/((z+2)*(z+1)z)

\int - \frac{1}{( z + 2 ) \cdot ( z + 1 ) z} d z

- \frac{1}{2} ln ∣ z + 2 ∣ + ln ∣ z + 1 ∣ - \frac{1}{2} ln ∣ z ∣ + C

Solution steps

\int - \frac{1}{( z + 2 ) ( z + 1 ) z} d z

Take the constant out:

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

= - \int \frac{1}{( z + 2 ) ( z + 1 ) z} d z

Take the partial fraction of

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

= - \int \frac{1}{2 ( z + 2 )} - \frac{1}{z + 1} + \frac{1}{2 z} d z

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 ( z + 2 )} d z - \int \frac{1}{z + 1} d z + \int \frac{1}{2 z} d z)

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

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

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

= - (\frac{1}{2} ln ∣ z + 2 ∣ - ln ∣ z + 1 ∣ + \frac{1}{2} ln ∣ z ∣)

Simplify

= - \frac{1}{2} ln ∣ z + 2 ∣ + ln ∣ z + 1 ∣ - \frac{1}{2} ln ∣ z ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ z + 2 ∣ + ln ∣ z + 1 ∣ - \frac{1}{2} ln ∣ z ∣ + C

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