The integral of the function 1/(x(x-1)) can be confirmed as -ln|x| + ln|x-1| + c using partial fractions. The method involves expanding the function into A/x + B/(x-1) to solve for constants A and B. Differentiating the resulting logarithmic expression verifies the correctness of the integral. This approach is essential for confirming the solution without redundancy.
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Can't you "confirm" it yourself by differentiating? The derivative of -ln |x|+ ln|x-1|+ cfrasifrasi 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.
I suspect that is exactly what he did! Doing it again is not a "check".Astronuc said:How about expanding [tex]\frac{1}{x (x-1)}[/tex] into
[tex]\frac{A}{x}\,+\,\frac{B}{x-1}[/tex] and solve for A and B
then solve the integral