Please confirm the answer for this integral

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Can anyone confirm that the answer
for int (1/x(x-1))dx

is

- ln|x|+ ln |x-1| + c using partial fractions?

Thank you.
 
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How about expanding \frac{1}{x (x-1)} into

\frac{A}{x}\,+\,\frac{B}{x-1} and solve for A and B

then solve the integral
 
frasifrasi said:
Can anyone confirm that the answer
for int (1/x(x-1))dx

is

- ln|x|+ ln |x-1| + c using partial fractions?

Thank you.
Can't you "confirm" it yourself by differentiating? The derivative of -ln |x|+ ln|x-1|+ c
(assuming for the moment that x> 1 so both x and x-1 are positive) is -1/x+ 1/(x-1)= -(x-1)/(x(x-1))+ x/(x(x-1))= 1(x(x-1)). (You might want to combine those two logarithms.)


Astronuc said:
How about expanding \frac{1}{x (x-1)} into

\frac{A}{x}\,+\,\frac{B}{x-1} and solve for A and B

then solve the integral
I suspect that is exactly what he did! Doing it again is not a "check".
 
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