What Techniques Can Simplify Integrating 1/(x^3(1-x^2))?

[avocadogirl]

Homework Statement

4[tex]\int[/tex] 1/ ( x^3 *( 1 - x^2)) dx

Homework Equations

Maybe integration by parts?

The Attempt at a Solution

This is actually a small fragment of a very large, long problem but, I cannot for the life of me think how to integrate this with there being a product, essentially, of (1/x^3) and (1/(1-x^2)), in the integrand.

I can see that, if I write the (1/(1-x^2)) like (1/(1+(-x^2))), I could sub the arctan trig identity. But, what about that pesky (1/x^3)?

Thank you.

 

[Saladsamurai]

[tex]4\int \frac{1}{x^3*(1-x^2)}\, dx[/tex]

Is this what you mean? I am just having a little trouble reading the integral.

 

[Count Iblis]

Try partial fractions.

 

[rootX]

Partial fractions?

(oo took some time replying...)

 

[Saladsamurai]

Third time's a charm! Partials

 

[avocadogirl]

But, which is u and which is dv?

 

[qntty]

Decompose it into partial fractions.

edit: didn't refresh so I was beat to it

 

[Saladsamurai]

U and dv [itex]\neq[/itex] to partial fractions

Here: http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/partial.html

 

[qntty]

But, which is u and which is dv?

Not integration by parts, use partial fractions

You should get that
[tex]
\frac{1}{x^3(1-x^2)} = \frac{1}{x^3} +\frac{1}{x}- \frac{1}{2x-2}-\frac{1}{2x+2}
[/tex]

 

[Count Iblis]

Partial fraction expansion for lazy people: Expand around the singularities, add up all the singular terms of all the expansions.

1) Expansion around x = 0:


1/x^3 1/(1-x^2) = 1/x^3 [1 + x^2 + ...] = 1/x^3 + 1/x + nonsingular terms


2) Expansion around x = ±1: We don't actually need to expand anything as we only need to find the coefficient of 1/(x -/+ 1) which we can do by multiplying by (x -/+ 1) and applying L'Hôpital's rule:

Lim x to ±1 of (x -/+ 1) 1/x^3 1/(1-x^2) = -1/2

The sum of all the singular terms of all the expansions is thus:


1/x^3 + 1/x - 1/2 [1/(x-1) + 1/(x+1)]

 

[Saladsamurai]

Not integration by parts, use partial fractions

You should get that
[tex]
\frac{1}{x^3(1-x^2)} = \frac{1}{x^3} - \frac{1}{2x-2}+\frac{1}{x}-\frac{1}{2 (1+x)}
[/tex]

We generally don't dole out the answers qntty. If you noticed the post before yours, I gave her a guide to solving them.

We got to let her a little fun

edit: just saw the Count's post too! I guess we do dole out answers!

 

[Count Iblis]

More explanation:

You can do a partial fraction expansion in the regular way by putting:


1/x^3 1/(1-x^2) = A/x^3 + B/x^2 + C/x + D/(x-1) + E/(x+1)

If you multiply both sides by x^3 (1-x^2), you get polynomials on both sides (the left hand side is just 1, of course). You can then equate te coeffiecent of all powers of x and then you get equations for the coefficients which you can solve.

This method is generally known as the "stupid high school method"


If we use some intelligent thinking, we can simplify things considerably. Instead of equating the coefficents of each power of x, you could choose certain values for x and then demand that both sides are equal. The clever values to choose are those values for which many terms vanish. If you choose x = 1, then only one term survives and you can then solve for D, if you choose x = -1, you get a single equation for E.

But this method does not allow you to get single equations for A, B, and C separately, so, it doesn't get rid of the need to solve simultaneous equations completely.


Sometimes you can use symmetries to simplify the problem. E.g. in this case the function is an odd function of x (i.e. it changes sign if you change the sign of x). This then implies that B = 0 and that the last to terms transform into each otther up to a minus sign. That then implies that D = E.


You can simplify things a lot more by proceeding in the following way. You can ask yourself what would happen if you take your rational function and would subtract from it all the singular terms in the expansion around the singular points. Then what you are left with doesn't have any singularities anymore, and must therefore be a polynomial. But if the degree of the numerator of the rational functon is less than the degree of the numerator the rational function tends to zero at infinity and so do the singular terms, so that difference must be zero. You must, of course, consider all the singular points in the complex plane.


So, you do a series expansion around x = 0, around x = 1 and
around x = -1, add up all the singular terms and then you got your partial fractions expansion. Since we're dealing with fractions, you only need to use the formula for te geometric series. In case the expansion starts at 1/(x-a) (e.g. in this case we have terms that will start with 1/(x-1)). yu only need to find the coefficient of that term. Then you can multiply the functon by x-a and take the limit for x to a. This is the equivalent of putting x = a in the undetermined constants method above.

Exercise for you:

Integrate 1/(x^6 + 1)