# Simple question for integral

• April 9th 2010, 12:19 AM
the prince
Simple question for integral
$\int \frac{dx}{x^4-1}$

$\frac{1}{x^4-1}= \frac{1}{(x-1)(x+1)(x^2+1)}$

$\frac{1}{x^4-1}=\frac{A}{(x-1)} +\frac{B}{x+1}+\frac{Cx+D}{x^2+1}$

why should i write Cx+D ,why not only C?
• April 9th 2010, 12:26 AM
Prove It
Quote:

Originally Posted by the prince
$\int \frac{dx}{x^4-1}$

$\frac{1}{x^4-1}= \frac{1}{(x-1)(x+1)(x^2+1)}$

$\frac{1}{x^4-1}=\frac{A}{(x-1)} +\frac{B}{x+1}+\frac{Cx+D}{x^2+1}$

why should i write Cx+D ,why not only C?

Because by creating partial fractions, you expect that the each numerator is of a smaller degree than the denominator.

If $x^2 + 1$, a degree 2 polynomial is the denominator, you would expect that the numerator would be a polynomial of smaller degree.

The largest possible polynomial is therefore a linear function (degree 1).