# Integration problem (partial fractions)

• November 10th 2010, 04:00 AM
SyNtHeSiS
Integration problem (partial fractions)
Integrate:

$\int \frac{2x^5 - 162x + 3}{x^4 - 81}dx$

Now I used partial fractions:

$\frac{2x^5 - 162x + 3}{x^4 - 81} = \frac{A}{x - 3} + \frac{B}{x + 3} + \frac{Dx + E}{x^2 + 9}$

and after a lot of working out got the following equations:

$A + B + D = 0$

$3A - 3B + E = 12$

$9A + 9B - 9D = -162$

$
27A - 27B - 9E = 3$

Now I tried gauss reducing it, but it was rather long and tedious which gave me:

$A = \frac{-167}{36}$

$B = \frac{157}{-36}$

$D = 9$

$E = -\frac{1}{6}$

In my final answer I got:

$\frac{-167}{36}ln|x - 3| - \frac{157}{36}ln|x + 3| + \frac{9}{2}ln|x^2 + 9| - \frac{1}{54}arctan(\frac{x}{3}) + C$

which was of that right form (due to the logs and arctan) but obviously the values were wrong.

So I was wondering, is there a more efficient way to do this question and solve for A, B, D and E without using gauss reduction?
• November 10th 2010, 05:38 AM
Unknown008
Well, I don't know if you can use that, but I use some sort of inspection to get the values of A and B.

$\dfrac{2x^5 - 162x + 3}{x^4 - 81} = \dfrac{A}{x - 3} + \dfrac{B}{x + 3} + \dfrac{Dx + E}{x^2 + 9}$

Take the fraction:

$\dfrac{2x^5 - 162x + 3}{(x+3)(x-3)(x^2 + 9)}$

Now, what value do you have to assign to x so that one denominator becomes 0?

We put x = 3.

So, you find this:

$\dfrac{2x^5 - 162x + 3}{(x+3)(x^2 + 9)}$

And put x = 3, to get 1/36. This is the value of A.

From this, take x = -3 now:

$\dfrac{2x^5 - 162x + 3}{(x-3)(x^2 + 9)}$

Evaluate and you get -1/36, which is the value of B.

Now, you are left with:

$\dfrac{2x^5 - 162x + 3}{(x+3)(x-3)(x^2+9)} = \dfrac{1}{36(x - 3)} - \dfrac{1}{36(x + 3)} + \dfrac{Dx + E}{x^2 + 9}$

There are only two constants to find now (Smile)
• November 10th 2010, 05:56 AM
SyNtHeSiS
Quote:

Originally Posted by Unknown008
Now, what value do you have to assign to x so that one denominator becomes 0?

We put x = 3.

So, you find this:

$\dfrac{2x^5 - 162x + 3}{(x+3)(x^2 + 9)}$

And put x = 3, to get 1/36. This is the value of A.

but then if you let x = 3, then wouldnt the whole of the LHS = 0? How would this get you a value for A, if you saying $\frac{A}{x - 3} = \frac{A}{3 - 3} = 0$?
• November 10th 2010, 06:04 AM
harish21
look at it this way:

$\dfrac{2x^5 - 162x + 3}{(x-3)(x+3)(x^2+9)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 3} + \dfrac{Dx + E}{x^2 + 9}$

So,

$2x^5 - 162x + 3 = A (x+3)(x^2+9) + B(x-3)(x^2+9) + (Dx+E)(x+3)(x-3)$..............................(I)

Now first put x=-3 in equation (I) to find A.

then put x=3 in equation (I) to find B..

Then work on to find D and E
• November 10th 2010, 06:21 AM
Unknown008
Yes, that's the same concept, but I find it faster as you don't have to write the full expansion before finding the values of A and B.
• November 10th 2010, 06:33 AM
harish21
Quote:

Originally Posted by Unknown008
Yes, that's the same concept, but I find it faster as you don't have to write the full expansion before finding the values of A and B.

Thats right! I thought the OP wasn't able to figure out the expansion before finding the values of A and B. So I wrote it down!!