1. 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

2. 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.

3. Originally Posted by Thomas154321
You copied it wrong! There should be no denominator by design.
what do you mean "by design"?

4. 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?

5. 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:
$\displaystyle ln(x) + ln(y) = ln(xy)$
$\displaystyle xln(y) = ln(y^x)$