How do you use partial fraction decomposition to decompose the fraction to integrate $(z+1)/(z^2(z^2+4))$ ?

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Answer 1 · 2018-07-04T00:43:34

$1/8 ln (z^2/(z^2+4))-1/(4z)-1/8tan^-1(z/2)+C$

While one could use the standard approach of beginning with

$(z+1)/(z^2(z^2+4)) = A/z + B/z^2+(Cz+D)/(z^2+4)$

it will be faster in this case to note that

$1/(z^2(z^2+4)) = 1/4(1/z^2-1/(z^2+4))$

and thus

$(z+1)/(z^2(z^2+4)) = (z+1)/4(1/z^2-1/(z^2+4))$

$qquadqquad =1/4((z+1)/z^2-(z+1)/(z^2+4))$

$qquadqquad = 1/4(1/z+1/z^2-(z+1)/(z^2+4))$

Thus

$int (z+1)/(z^2(z^2+4)) dz = int 1/4(1/z+1/z^2-(z+1)/(z^2+4))dz$

$qquad = 1/4 log z-1/(4z) - 1/8 ln (z^2+4)-1/8 tan^-1(z/2)+C$

$qquad= 1/8 ln (z^2/(z^2+4))-1/(4z)-1/8tan^-1(z/2)+C$

How do you use partial fraction decomposition to decompose the fraction to integrate (z+1)/(z^2(z^2+4))(z+1)/(z^2(z^2+4))?