• Support PF! Buy your school textbooks, materials and every day products Here!

Algebraic Integral

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


[itex] \displaystyle \int_0^{\infty} \dfrac{x^2-1}{x^4+1} dx [/itex]

Homework Equations



The Attempt at a Solution


If I separate the integral I get

[itex] \displaystyle \int_0^{\infty} \dfrac{x^2}{x^4+1} dx - \int_0^{\infty} \dfrac{1}{x^4+1} dx[/itex]

But these two integrals are itself complicated.
 

Answers and Replies

  • #2
329
34
Yes, it is a difficult integral. Here's an approach using partial fractions. I may not have written it quite right -- please check the arithmetic.

Let's start with ## x^4 + 1## . This can be rewritten as ##x^4 + 2x^2 +1 - 2x^2 = (x^2 + 1)^2 - 2x^2##. You can factor this expression as ##(x^2 +1 - \sqrt 2x)(x^2 + 1 + \sqrt2x)##. You use the partial fraction method to rewrite this as
##\frac{x^2-1}{(x^2 +1 - \sqrt 2x)(x^2 + 1 + \sqrt2x)} = \frac{A}{(x^2 +1 - \sqrt 2x)} + \frac{B}{x^2 + 1 + \sqrt2x}## where A and B are real numbers.

Now add up these fractions exactly as you did in elementary school and set the numerator to ##x^2 - 1##. This will allow you to solve for A and B.

Now how do you integrate ## \frac{A}{(x^2 +1 - \sqrt 2x)}##? This one you can look up. ##\int \frac{1}{ax^2 + bx + c}dx = \frac{2}{d}arctan \frac{2ax + b}{d}## where d = ##\sqrt{4ac - b^2}##
 
  • #3
3,812
92

Homework Statement


[itex] \displaystyle \int_0^{\infty} \dfrac{x^2-1}{x^4+1} dx [/itex]

Homework Equations



The Attempt at a Solution


If I separate the integral I get

[itex] \displaystyle \int_0^{\infty} \dfrac{x^2}{x^4+1} dx - \int_0^{\infty} \dfrac{1}{x^4+1} dx[/itex]

But these two integrals are itself complicated.
Rewrite the given integral as:
[tex]\int_0^{\infty} \frac{1-1/x^2}{x^2+1/x^2} dx[/tex]

##1-1/x^2## is the derivative of ##x+1/x##. Do you see how to proceed from here?
 
  • Like
Likes 1 person
  • #4
329
34
Rewrite the given integral as:
[tex]\int_0^{\infty} \frac{1-1/x^2}{x^2+1/x^2} dx[/tex]

##1-1/x^2## is the derivative of ##x+1/x##. Do you see how to proceed from here?
Very clever, I like it.
 

Related Threads on Algebraic Integral

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
603
Replies
3
Views
359
Replies
5
Views
4K
Replies
4
Views
676
Replies
10
Views
360
Replies
1
Views
2K
Replies
4
Views
981
Replies
6
Views
1K
Top