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

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

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

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

- \frac{( z - 1 ) ^{2}}{z ^{3}} - \frac{1}{z ^{2}} + \frac{2}{3 z ^{3}} + C

Solution steps

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

Apply Integration By Parts

: - \frac{( z - 1 ) ^{2}}{z ^{3}} - \int - \frac{2 ( z - 1 )}{z ^{4}} d z

= - \frac{( z - 1 ) ^{2}}{z ^{3}} - \int - \frac{2 ( z - 1 )}{z ^{4}} d z

\int - \frac{2 ( z - 1 )}{z ^{4}} d z = \frac{1}{z ^{2}} - \frac{2}{3 z ^{3}}

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

Simplify

= - \frac{( z - 1 ) ^{2}}{z ^{3}} - \frac{1}{z ^{2}} + \frac{2}{3 z ^{3}}

Add a constant to the solution

= - \frac{( z - 1 ) ^{2}}{z ^{3}} - \frac{1}{z ^{2}} + \frac{2}{3 z ^{3}} + C

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