What is the integral of (4z^3)/(z^2-z^3) ?

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

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integral of (4z^3)/(z^2-z^3)

\int \frac{4 z ^{3}}{z ^{2} - z ^{3}} d z

4 (\frac{z ^{2}}{1 - z} - ln ∣ z - 1 ∣ + \frac{1}{z - 1}) + C

Solution steps

\int \frac{4 z ^{3}}{z ^{2} - z ^{3}} d z

Take the constant out:

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

= 4 \cdot \int \frac{z ^{3}}{z ^{2} - z ^{3}} d z

Apply Integration By Parts

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

= 4 (\frac{z ^{2}}{1 - z} - \int \frac{z}{( z - 1 ) ^{2}} d z)

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

= 4 (\frac{z ^{2}}{1 - z} - (ln ∣ z - 1 ∣ - \frac{1}{z - 1}))

Simplify

= 4 (\frac{z ^{2}}{1 - z} - ln ∣ z - 1 ∣ + \frac{1}{z - 1})

Add a constant to the solution

= 4 (\frac{z ^{2}}{1 - z} - ln ∣ z - 1 ∣ + \frac{1}{z - 1}) + C

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