Partial Fraction Question

[cmhex]

EDIT: nvm, we got it, no option to delete post or I would.

So its an integral 1 to 2

4y^2-6x-12 / y(y+2)(y-3)

I have the answer all the way down to

24/5 ln(2) - 8/5 ln(3)

Now here's the somewhat embarrassing part.

How do get from

26/5 ln(2) - 8/5ln(3)

to

8/5 ln (8/3)

I know the 8/5 gets pulled out making it 8/5 (ln(8)-ln(3)), but its not clear to me where that ln(8) came from.

 

[HallsofIvy]

EDIT: nvm, we got it, no option to delete post or I would.

It's better not to delete. Others can learn from your post.

So its an integral 1 to 2

4y^2-6x-12 / y(y+2)(y-3)

$\displaystyle \int_1^2 \frac{4y^2- 6y- 12}{y(y+ 2)(y- 3)}dy$
You have "6x" but I assume it was supposed to be "6y".

It is possible that we can cancel something so check if y+ 2 or y- 3, factors in the denominator, are factors of the numerator: $\displaystyle 4(-2)^2- 6(-2)- 12= 16+ 12- 12= 16$ not 0 so y+ 2 is not a factor of the numerator. $\displaystyle 4(3)^2- 6(3)- 12= 36- 18- 12= 6.$No, y- 3 is not a factor of the numerator.

So we use "partial fractions". Find numbers A, B, and C such that $\displaystyle \frac{4y^2- 6y- 12}{y(y+ 2)(y- 3)}= \frac{A}{y}+ \frac{B}{y+ 2}+ \frac{C}{y- 3}$. Once we have done that, the integral will be of the form $\displaystyle \displaystyle {\left[A ln|y|+ B ln|y+ 2|+ C ln|y-3|\right]_1^2}= A(ln 3- ln 2)+ B(ln 4- ln 3)+ C(ln 1- ln 2)$

I have the answer all the way down to

24/5 ln(2) - 8/5 ln(3)

Now here's the somewhat embarrassing part.

How do get from

26/5 ln(2) - 8/5ln(3)

Is it "24/5", as before, or "26/5"? 24= 3(8) so if it is really (24/5)ln(2)- (8/5) ln(3) then it is (8/5)( 3 ln(2)- ln(3))= (8/5)(ln(2^3)- ln(3))= (8/5)(ln(8)- ln(3))= (8/5) ln(8/3)

to

8/5 ln (8/3)

I know the 8/5 gets pulled out making it 8/5 (ln(8)-ln(3)), but its not clear to me where that ln(8) came from.

2^3= 8.