Question

kindly help me integrate this please.

zdz/(z^2+4z+4)

i think i have a partial fraction here.

Answer 1

\[\int \frac{z\cdot\text dz}{z^2+4z+4}\]

Answer 2

\[\frac{z}{z^2+4z+4}=\frac z{(z+2)^2}=\frac A{z+2}+\frac B{(z+2)^2}\]

Answer 3

{z/(z+2)(z+2)}
{z/(z+2)^2}
1/z+2 - 2/(z+2)^2 integrating
ln(z+2) +2/z+2 i think that i have done it right

Answer 4

\[z=(z+2)A+B\]
\[z=0\Rightarrow0=2A+B\]
\[\qquad\qquad\qquad z=-2 \Rightarrow -2=B\]
\[0=2A-2\rightarrow 1=A\]


\[\frac{z}{z^2+4z+4}=\frac z{(z+2)^2}=\frac 1{z+2}-\frac 2{(z+2)^2}\]

\[\int\frac1{z+2}-\frac 2{(z+2)^2}\text dz\]

Answer 5

\[\int\frac1{z+2}-\frac 2{(z+2)^2}\text dz\]
\[=\ln(z+2)+\frac2{z+2}+c\]

Answer 6

\[\int\frac{z}{z^2+4z+4}dz\]
\[\int\frac{z+(2-2)}{z^2+4z+4}dz\]
\[\int\frac{2}{2}*\frac{z+2-2}{z^2+4z+4}dz\]
\[\frac{1}{2}\int\frac{2z+4-4}{z^2+4z+4}dz\]
\[\frac{1}{2}\int\frac{2z+4}{z^2+4z+4}dz-\frac42\int\frac{1}{(z+2)^2}dz\]
\[ln(z+2)+2 (z+2)^{-1}+C\]