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

Printable View

- Apr 6th 2009, 05:41 AMur5pointos2sloPartial 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 - Apr 6th 2009, 06:05 AMThomas154321
- Apr 6th 2009, 06:12 AMur5pointos2slo
- Apr 6th 2009, 06:25 AMur5pointos2slo
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? - Apr 6th 2009, 01:18 PMThomas154321
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: