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Are you differentiating [itex]\ln(u^2 -u)[/itex] wrt "x" or to "u" ?
Why would you think that?camilus said:[tex]\int{1 \over u^2-u}du = ln(u^2-u) + C[/tex]
Hurkyl said:Solve a man's integral problem, and he gets one answer. Teach a man how to solve integral problems, and he gets all of his homework done.(and without cheating)