Find the integral of: x^3/(x^2 + 4)(x^2 - 4) dx

[MarkSA]

Hello,

1) Find the integral of x^3/(x^2 + 4)(x^2 - 4) dx

I am having some trouble with these types of problems. This one requires partial fraction decomposition I think.

I multiplied the denominator out which gives:
x^3/(x^2 + 4)(x + 2)(x - 2)

And I tried to decompose that...

x^3/[(x^2 + 4)(x + 2)(x - 2)] = (Ax + B)/(x^2 + 4) + C/(x + 2) + D(x - 2)

I must be doing something wrong. I multiplied that whole mess out by the denominator on the left and ended up with an algebraic expression the entire length of my page. Worst was that on the left I only had x^3. So, if I group terms, I think all I ended up with was x^3(A + B + D) and everything else x^2, x, and constants were all equal to 0. All I really had was: A + B + D = 1. I couldn't figure out a way to find the values of the variables with just that.

Do you know where i'm going wrong at?

 

[stapel]

I multiplied the denominator out...

I'm guessing you mean that you did the opposite, and factored the denominator...?

I must be doing something wrong....

Do you know where i'm going wrong at?

Since we can't see your work, no, I'm afraid we cannot know wherein lies the error. Sorry.

Using your starting point and multiplying through on both sides, I end up (after three lines of work) with the following equation:

. . . . .\(\displaystyle (A\, +\, C\, +\, D)x^3\, +\, (B\, -\, 2C\, +2D)x^2\, +\, (-4A\, +\, 4C\, -\, 4D)x\, +\, (-4B\, -\, 8C\, +\, 8D)1\)
. . . . . . . . . . .\(\displaystyle =\, (1)x^3\, +\, (0)x^2\, +\, (0)x\, +\, (0)1\)

This creates the following system of equations:

. . . . .\(\displaystyle \begin{array}{ccccccccc}A&&&+&C&+&D&=&1 \\ &&B&-&2C&+&2D&=&0 \\ -4A&&&+&4C&+&4D&=&0 \\ &-&4B&-&8C&+&8D&=&0 \end{array}\)

This should be solvable. :wink:

Eliz.

 

[MarkSA]

Is there some easier or quicker way to solve this problem than doing it that way?

I ask because we had this question on a quiz and it was taking an enormous amount of time doing it this way. I'm thinking there must have been a better way since I ran out of time on this one.

 

[galactus]

Yes, there is a quicker way.

It's called a TI-89, TI-92, Voyage 200, or other nice calculator. I am with pka on this anachronistic partial fractions business.

But...here is a technique that sometimes makes it easier. I will illustrate with an example. See if you can apply it to yours.

\(\displaystyle \frac{x^{2}+4x-1}{(x-1)(x-2)(x+3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+3}\)

Now, suppose we multiply both sides of the last expresion by x-1, simplify, and then set x = 1.
Since the coefficients of B and C are 0 we get:

\(\displaystyle A = \frac{(x^{2}+4x-1}{(x-2)(x+3)}=\frac{1^{2}+4(1)-1}{(1-2)(1+3)} = -1\)

Write another way:

\(\displaystyle \frac{x^{2}+4x-1}{\boxed{(x-1)}(x-2)(x+3)}=A\)

Now do B and C by using x-2 and x+3. So, sub in x=2 and x=-3.

We get \(\displaystyle B=\frac{11}{5}\) and \(\displaystyle C=\frac{-1}{5}\)

So, we get:

\(\displaystyle \frac{-1}{x-1}+\frac{11}{5(x-2)}-\frac{1}{5(x+3)}\)

This is called the cover up method. It is a simplified version of the Heaviside Expansion Thereom.

Give it a try. I like to use it, instead of equating coefficients, on PFD if I ever have to do them

 

[MarkSA]

Yes, there is a quicker way.

It's called a TI-89,

Haha!

I wish we could use it but they've decreed no calculators for this chapter.. but, I feel a little better now for not finishing this problem.

Thanks for the reply & the short cut, I will try to understand it.

 

[galactus]

Here's your PFD. Gives you something to shoot for. My TI-92 gave it to me in one second. That's why I agree with pka on this.
Along with Gaussian elimination.

\(\displaystyle \frac{x}{2(x^{2}+4)}+\frac{1}{4(x+2)}+\frac{1}{4(x-2)}\)

Of course, you probably realize your integrals will be in terms of the natural log.

 

[Deleted member 4993]

Hello,

1) Find the integral of x^3/(x^2 + 4)(x^2 - 4) dx

I am having some trouble with these types of problems. This one requires partial fraction decomposition I think.

I multiplied the denominator out which gives:
x^3/(x^2 + 4)(x + 2)(x - 2)

And I tried to decompose that...

x^3/[(x^2 + 4)(x + 2)(x - 2)] = (Ax + B)/(x^2 + 4) + C/(x + 2) + D(x - 2)

I must be doing something wrong. I multiplied that whole mess out by the denominator on the left and ended up with an algebraic expression the entire length of my page. Worst was that on the left I only had x^3. So, if I group terms, I think all I ended up with was x^3(A + B + D) and everything else x^2, x, and constants were all equal to 0. All I really had was: A + B + D = 1. I couldn't figure out a way to find the values of the variables with just that.

Do you know where i'm going wrong at?

The quickest way to do "this" integral:

\(\displaystyle \int{\frac{x^3}{(x^2+4)(x^2-4)}}dx\)

\(\displaystyle \int{\frac{x^3}{(x^4 - 16)}}dx\)

substitute

\(\displaystyle u = x^4 - 16\)

\(\displaystyle du = 4\cdot x^3\, dx\)

\(\displaystyle \int{\frac{x^3}{(x^4 - 16)}}dx\)

\(\displaystyle = \frac{1}{4}\int{\frac{du}{u}}\)

Now finish it

Do not get hooked on PF whenever you see a fraction.

 

[stapel]

The quickest way to do "this" integral:

\(\displaystyle \int{\frac{x^3}{(x^2+4)(x^2-4)}}dx\)

\(\displaystyle \int{\frac{x^3}{(x^4 - 16)}}dx\)

substitute \(\displaystyle u = x^4 - 16\) ...

Oo! That's very nice!

Eliz.

 

[MarkSA]

Oh no! I can't believe I fell into that one.

I actually had it rewritten out in the form of x^3/(x^4 - 16) but didn't think to substitute. Tricky!