# Algebraic Integral

1. Oct 10, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
$\displaystyle \int_0^{\infty} \dfrac{x^2-1}{x^4+1} dx$

2. Relevant equations

3. The attempt at a solution
If I separate the integral I get

$\displaystyle \int_0^{\infty} \dfrac{x^2}{x^4+1} dx - \int_0^{\infty} \dfrac{1}{x^4+1} dx$

But these two integrals are itself complicated.

2. Oct 11, 2013

### brmath

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. Oct 11, 2013

### Saitama

Rewrite the given integral as:
$$\int_0^{\infty} \frac{1-1/x^2}{x^2+1/x^2} dx$$

$1-1/x^2$ is the derivative of $x+1/x$. Do you see how to proceed from here?

4. Oct 11, 2013

### brmath

Very clever, I like it.