Question: How to perform this complex substitution

 

Some more difficult calculation with complex numbers

Starting from the equation     
A = arctan(z/b)/b and arctan(z/b)/b = int(1/(b^2+z^2), z = 0 .. z)
made the substitution                               
z = i*b(t-1)/(t+1)
Thus obtaining NULL
A = -(int(1/t, t = 1 .. t))/(2*bi) and -(int(1/t, t = 1 .. t))/(2*bi) = log[10](t)/(2*bi) and log[10](t)/(2*bi) = log[10]((b*i+z)/(b*i-z))/(2*bi)
One of the important relations between logarithms and inverse trigonometrische functions

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