Page:Calculus Made Easy.pdf/147

This method can always be used; but the method shown first will be found the quickest in the case of factors in $x$ only.

Case III. When, among the factors of the denominator there are some which are raised to some power, one must allow for the possible existence of partial fractions having for denominator the several powers of that factor up to the highest. For instance, in splitting the fraction ${\dfrac {3x^{2}-2x+1}{(x+1)^{2}(x-2)}}$ we must allow for the possible existence of a denominator $x+1$ as well as $(x+1)^{2}$ and $(x-2)$.

It maybe thought, however, that, since the numerator of the fraction the denominator of which is $(x+1)^{2}$ may contain terms in $x$, we must allow for this in writing $Ax+B$ for its numerator, so that

${\frac {3x^{2}-2x+1}{(x+1)^{2}(x-2)}}={\frac {Ax+B}{(x+1)^{2}}}+{\frac {C}{x+1}}+{\frac {D}{x-2}}$.

If, however, we try to find $A$, $B$, $C$ and $D$ in this case, we fail, because we get four unknowns; and we have only three relations connecting them, yet

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

But if we write

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

we get

$3x^{2}-2x+1=A(x-2)+B(x+1)(x-2)+C(x+1)^{2}$,