$\displaystyle \int \frac{dx}{x^4-1}$

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

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

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

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- Apr 8th 2010, 11:19 PMthe princeSimple question for integral
$\displaystyle \int \frac{dx}{x^4-1}$

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

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

why should i write Cx+D ,why not only C? - Apr 8th 2010, 11:26 PMProve It
Because by creating partial fractions, you expect that the each numerator is of a smaller degree than the denominator.

If $\displaystyle x^2 + 1$, 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).