Page:Calculus Made Easy.pdf/143

Suppose we wish to go back from ${\dfrac {3x+1}{x^{2}-1}}$ to the components which we know are ${\dfrac {1}{x+1}}$ and ${\dfrac {2}{x-1}}$. If we did not know what those components were we can still prepare the way by writing:

${\frac {3x+1}{x^{2}-1}}={\frac {3x+1}{(x+1)(x-1)}}={\frac {}{x+1}}+{\frac {}{x-1}}$,

leaving blank the places for the numerators until we know what to put there. We always may assume the sign between the partial fractions to be plus, since, if it be minus, we shall simply find the corresponding numerator to be negative. Now, since the partial fractions are proper fractions, the numerators are mere numbers without $x$ at all, and we can call them $A$, $B$, $C$ . . . as we please. So, in this case, we have:

${\frac {3x+1}{x^{2}-1}}={\frac {A}{x+1}}+{\frac {B}{x-1}}$.

If now we perform the addition of these two partial fractions, we get ${\dfrac {A(x-1)+B(x+1)}{(x+1)(x-1)}}$; and this must be equal to ${\dfrac {3x+1}{(x+1)(x-1)}}$. And, as the denominators in these two expressions are the same, the numerators must be equal, giving us:

$3x+1=A(x-1)+B(x+1)$.

Now, this is an equation with two unknown quantities, and it would seem that we need another equation before we can solve them and find $A$ and $B$.