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

Answer

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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

$\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

Last updated date: 26th Sep 2023

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