Question

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

Answer 1

\[\int \frac{z^2}{(z-1)^3} dz\]
Substitute z=u+1
so
\[dz=du\]
We have now
\[\int \frac{(u+1)^2}{(u)^3} dz\]
Let's expand the square term
\[\int \frac{(u^2+1+2u)}{(u)^3} dz\]
we get
\[\int \frac{1}{u}+ \frac{1}{u^3}+\frac{2}{u^2} du\]
Can you do it now?

Answer 2

but there is another way which i know and that's in which we do something like in numerator we place constants like A B C......

Answer 3

That's the method of Partial fractions. Did you try it?

Answer 4

the one which you have mentioned i dont used to solve partial fraction question...i used to take that method which i m telling you

Answer 5

z^2= A/(z-1)^2 + B(z-1)

Answer 6

this method will work, but lord is it a pain

Answer 7

you will need
\[\frac{z^2}{(z-1)^3}=\frac{a}{z-1}+\frac{b}{(z-1)^2}+\frac{c}{(z-1)^3}\] to do this

Answer 8

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