What is the integral of (z^2)/((z-1)^3) ?

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

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integral of (z^2)/((z-1)^3)

\int \frac{z ^{2}}{( z - 1 ) ^{3}} d z

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

Solution steps

\int \frac{z ^{2}}{( z - 1 ) ^{3}} d z

Apply u-substitution

: \int \frac{( u + 1 ) ^{2}}{u ^{3}} d u

= \int \frac{( u + 1 ) ^{2}}{u ^{3}} d u

Expand

\frac{( u + 1 ) ^{2}}{u ^{3}} : \frac{1}{u} + \frac{2}{u ^{2}} + \frac{1}{u ^{3}}

= \int \frac{1}{u} + \frac{2}{u ^{2}} + \frac{1}{u ^{3}} d u

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}{u} d u + \int \frac{2}{u ^{2}} d u + \int \frac{1}{u ^{3}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{2}{u ^{2}} d u = - \frac{2}{u}

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= ln ∣ u ∣ - \frac{2}{u} - \frac{1}{2 u ^{2}}

Substitute back

u = z - 1

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

Add a constant to the solution

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

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