solving for simple Integration symbol

When integrate for rho in this equation b/(1-b*rho), I get -log(b*rho-1) which is wrong. It should come out to be -log(1-b*rho).

My code:

syms A B Z a b R T rho

q1 = b/(1-b*rho)

I1 = int(q1,rho)

I1 = - log(b*rho - 1)

Answer 1

[David Goodmanson]

HI AC

d/drho (-)*log(b*rho-1) = (-)*1/(b*rho-1)*b = b/(1-b*rho) = q1

so it is correct. But your result is correct as well.

Ignoring the (-) in front for the moment, your result is

log(1-b*rho) = log((-1)*(b*rho-1)) = log(b*rho-1) + log(-1)
= log(b*rho-1) +i*pi

which differs from the 'int' result by a constant of integration i*pi. Whichever result you want to use would usually be decided by keeping the argument of the log function to be positive.