What are the Coefficients for Partial Fractions?

[Yukz23]

I have trouble solving the following integral $\frac{x-6}{x^2-4}$

If I let u= x^2 - 4, I'm left with 1/2du = x dx

and I'm stuck trying to get rid of that 6 in the numerator?

If someone can help me out, that would be good.

 

[haruspex]

Do you know how to rewrite expressions like 1/((ax+b)(cx+d)) as a sum of simpler fractions?

 

[SithsNGiggles]

There's another method besides the one that @haruspex mentioned. You can also rewrite the numerator so that it contains a du.

For example,
$\frac{x + 2}{\frac{1}{2}x^2 + 3x}$

Letting
$u = \frac{1}{2}x^2 + 3x$,

$du = x + 3$.

Then the original numerator can be manipulated algebraically to get a proper substitution involving du, plus an additional term.

Sticking to my example,
$\frac{x + 2}{\frac{1}{2}x^2 + 3x}$

$\frac{x + 3 - 1}{\frac{1}{2}x^2 + 3x}$

$\frac{(du) - 1}{u}$

 

[haruspex]

$du = x + 3$.

You mean $du = (x + 3)dx$

$\frac{x + 2}{\frac{1}{2}x^2 + 3x}$
$\frac{x + 3 - 1}{\frac{1}{2}x^2 + 3x}$
$\frac{(du) - 1}{u}$

Filling in the details exposes a flaw:
$\frac{x + 2}{\frac{1}{2}x^2 + 3x}dx$
$\frac{x + 3 - 1}{\frac{1}{2}x^2 + 3x}dx$
$\frac{du - dx}{u}$

 

[SammyS]

There's another method besides the one that @haruspex mentioned. You can also rewrite the numerator so that it contains a du.

For example,
$\frac{x + 2}{\frac{1}{2}x^2 + 3x}$

Letting
$u = \frac{1}{2}x^2 + 3x$,

$du = x + 3$.

Actually, $du = (x + 3)dx$

Then the original numerator can be manipulated algebraically to get a proper substitution involving du, plus an additional term.

Sticking to my example,
$\frac{x + 2}{\frac{1}{2}x^2 + 3x}$

$\frac{x + 3 - 1}{\frac{1}{2}x^2 + 3x}$

$\frac{(du) - 1}{u}$

You end up with only being able to use u for part of the integrand.

$\displaystyle \frac{x + 2}{\frac{1}{2}x^2 + 3x}\quad\to\quad\frac{(du) - dx}{u} \quad\to\quad \frac{du}{u}-\frac{dx}{\frac{1}{2}x^2 + 3x}$

 

[Yukz23]

Why are you guys doing the part where its du - dx in the numerator I don't get where the dx part is coming from? and what's its value?

 

[haruspex]

Why are you guys doing the part where its du - dx in the numerator I don't get where the dx part is coming from? and what's its value?

This part of the thread was triggered by SithsNGiggles' post, which had an error in it. SammyS and I jumped into stop you being misled by it.
Please go back to my first post.

 

[SammyS]

Why are you guys doing the part where its du - dx in the numerator I don't get where the dx part is coming from? and what's its value?

I was merely responding to SithsNGiggles's post -- that's why I 'Quoted' it. His suggestion was erroneous.

What haruspex suggested is the way to go. That is, to use partial fraction decomposition to write $\displaystyle \frac{x-6}{x^2-4}$ as the sum of two fractions, one with a denominator of x-2 the other with a denominator of x+2 .

 

[Yukz23]

I was merely responding to SithsNGiggles's post -- that's why I 'Quoted' it. His suggestion was erroneous.

What haruspex suggested is the way to go. That is, to use partial fraction decomposition to write $\displaystyle \frac{x-6}{x^2-4}$ as the sum of two fractions, one with a denominator of x-2 the other with a denominator of x+2 .

ohh is that the integral where we do the A/x-2 + B/x+2 thing? I don't really remember doing that a lot I forgot what the values of A and B would be

 

[haruspex]

ohh is that the integral where we do the A/x-2 + B/x+2 thing? I don't really remember doing that a lot I forgot what the values of A and B would be

You don't have to remember. Just write down that your fraction equals A/(x-2) + B/(x+2), multiply out, and find the values of A and B that make all the coefficients match up.

 

[SithsNGiggles]

Oh, sorry about that. I forgot to show the step where the fraction was split to du/u + dx/u. Thanks for catching that.

 

[Yukz23]

ok guys thanks I solved this problem pretty sure I'm correct to,
i got
2ln|x+2| + lin|x-2| + C

hmm the solution says - lin|x-2| on the second part , why is it - rather than +

 

[Dick]

ok guys thanks I solved this problem pretty sure I'm correct to,
i got
2ln|x+2| + lin|x-2| + C

hmm the solution says - lin|x-2| on the second part , why is it - rather than +

What did you get for A and B in the partial fractions expansion? You should find your - sign there.

 

[Yukz23]

What did you get for A and B in the partial fractions expansion? You should find your - sign there.

I had
A+B=1
-2A + 2B= -6

multiplied top by 2 then substracted with bottom,
2A + 2B = 2
-2A +2B = -6

Subtracting the top with bottom I got
4A = 8
A = 2,
sub A into the first equation

2 + B = 1
2-1 = -B

-1 = B
OH WHOOPS I didnt make B negative when I moved it to the other side!