Integrate $\int {{{\left( {\frac{{1 - z}}{z}} \right)}^2}dz} $.

Hint- First, We simplify this equation by using of basic formula of algebra and integrate with respect to $z$
$\int {{{\left( {\frac{{1 - z}}{z}} \right)}^2}dz} $
$\int {\frac{{{{\left( {1 - z} \right)}^2}}}{{{z^2}}}dz} $
Here we use ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$\int {\frac{{\left( {1 + {z^2} - 2z} \right)}}{{{z^2}}}dz} $
Whole equation is divided by ${z^2}$
$\int {\left( {\frac{1}{{{z^2}}} + 1 - \frac{2}{z}} \right)dz} $
Now, Integrate with respect to $z$
$ - \frac{1}{z} + z - 2{\log _e}z + $$c$ (here $c$is an integral constant)
Ans.
Note- In this type of question first we simplify the equation after that apply basic formula of integral