Thread: Partial fraction

so factoring the bottom we get



so we have

A / x-1

B / x+6

now what? I know you have to do this

x+4 = A(x+6) + B(x-1)(x+6)/(x-6)

now im stuck because of the x-6 in denominator

Originally Posted by ur5pointos2slo

so we have

B / x+6

x+4 = A(x+6) + B(x-1)(x+6)/(x-6)

You copied it wrong! There should be no denominator by design.

Originally Posted by Thomas154321

You copied it wrong! There should be no denominator by design.

what do you mean "by design"?

ok so Ive tried to work it a little further.

a= 5/7
b=2/7

is this correct?

then I get



ok now I know this is a du/u but the book has this answer

1/7 ln|(x+6)^2(x-1)^5| + C

is there some log property they used to get to this form?

All I meant with "by design" was that the denominator will always cancel because of the way partial fractions work.



Those values for a and b are correct.



This is as you say du/u which integrates to ln(u).

So we have which is what you have I hope?

The following are true for logs: