How do you find #int (x - 4 ) / (x^2 -4x) dx# using partial fractions?

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Answer 1 · 2015-11-04T14:56:16

# int (x-4)/(x^2-4x)dx = ln |x| #

# int (x-4)/(x^2-4x)dx = int (x-4)/(x(x-4))dx = int 1/x dx = ln |x| #

Answer 2 · 2015-11-04T15:02:04

Reduce the fraction.

When you factor the denominator to start the partial fraction decomposition, note that the ratio can be reduced.

#int (x-4)/(x^2-4x) dx = int (x-4)/(x(x-4)) dx#

# = int 1/x dx = lnabsx +C#

If you didn't notice it could be reduced, find #A, "and " B# so that:

#A/x + B/(x-4) = (x-4)/(x(x-4))#

So #Ax-4A+Bx = x-4#

And #A+B=1#

and #-4A=-4#, #" "# so #A=1# and #B=0#