# Partial fraction

• Apr 6th 2009, 05:41 AM
ur5pointos2slo
Partial fraction
$\displaystyle \int\frac{x+4}{x^2+5x-6}\,dx$

so factoring the bottom we get

$\displaystyle \int\frac{x+4}{(x-1)(x+6)}\,dx$

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 AM
Thomas154321
Quote:

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.
• Apr 6th 2009, 06:12 AM
ur5pointos2slo
Quote:

Originally Posted by Thomas154321
You copied it wrong! There should be no denominator by design.

what do you mean "by design"?
• Apr 6th 2009, 06:25 AM
ur5pointos2slo
ok so Ive tried to work it a little further.

a= 5/7
b=2/7

is this correct?

then I get

$\displaystyle 1/7 \int\frac{5}{x-1}+\int\frac{2}{x+6}\,dx$

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 PM
Thomas154321
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.

$\displaystyle \frac{1}{7} \int\frac{5}{x-1}+\frac{2}{x+6}~dx$

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

So we have $\displaystyle \frac{1}{7}(5ln(x-1)+2ln(x+6))$ which is what you have I hope?

The following are true for logs:
$\displaystyle ln(x) + ln(y) = ln(xy)$
$\displaystyle xln(y) = ln(y^x)$