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

Question

1778 views around the world

You can reuse this answer
Creative Commons License

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#