# Algebraic Integral

Gold Member

## Homework Statement

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

## 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.

Related Calculus and Beyond Homework Help News on Phys.org
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}##

## Homework Statement

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

## 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.
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?

1 person
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?
Very clever, I like it.