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How do you determine the indefinite integral of #(9/(4+z^2))dz#?

1 Answer

May 9, 2015

In this way, remembering that

#int(f'(x))/(1+[f(x)]^2)dx=arctgf(x)+c#:

#int(9/(4+z^2))dz=9int1/(4(1+z^2/4))dz=9/4int1/(1+(z/2)^2)dz=#

#=2*9/4int(1/2)/(1+(z/2)^2)dz=9/2arctg(z/2)+c#.

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