Partial Fraction Decomposition

[jmedina94]

My professor asks us to solve the integral of:

[x/(x^4 + 1)]dx

This expression is not factorable; what should I do? She is asking us to solve specifically using PFD, not u-substitution.

 

[arildno]

It most certainly is factorable!
Are you acquainted with complex numbers? (they can be extremely handy on an intermediate step in the factorization procedure)

 

[jmedina94]

The only method I thought of doing would be to complete the square.

 

[arildno]

The only method I thought of doing would be to complete the square.

The problem with that is, as I'm sure you've noticed, you don't know what to add or subtract. Agreed?

So, we'll proceed differently:
1. ALL polynomials can be factored, with factors being polynomials of degree 2 or less.
2. Thus, the following equation MUST have a solution, for ALL x:
(x^{2}+ax+b)*(x^2+cx+d)=x^{4}+1 (**)
where a,b,c and d are constants to be determined somehow.
3. Now, precisely HOW can a,b,c and d be determined? We'll basically need 4 equations to do so!
Those four equations are gained as follows:
When you multiply together the polynomials on your left hand side LHS of (**), you'll get a term (a+c)*x^3
But, your RHS does NOT contain any terms of power x^3,
thus we must have a+c=0

Similarly, you'll get a term multiplied with x^2, the coefficient of that one must be zero also, by the above argument.

You also must have zero equal to the factor multiplied with x, yielding your third equation.
But finally, multiplying b*d, we see from the RHS in (**) that it must equal 1, that is your fourth and final equation to determine a,b,c and d must be b*d=1
------------------
4. Now, set up all the four equations I have indicated to you; you may try to solve them (it's not that difficult, actually!), or return for more help!

5. Have you understood what you are to do, and why this is a constructive approach?

 

[arildno]

I'm sorry!

OF COURSE you can solve this simpler by completing the square!

Rewrite this as:
(x^{4}+2x^{2}+1)-(\sqrt{2}x)^{2}
Now, you first write the first parenthesis as a square; then use the identity a^2-b^2=(a+b)(a-b) in order to complete factorization.

 

[jmedina94]

How could I factor it?

 

[arildno]

Have you read post 5?

 

[jmedina94]

Have you read post 5?

I'm sorry, I'm on my mobile phone :-) thank you, I am reading it right now and will post back with an answer.