Page:Calculus Made Easy.pdf/148

which gives $C=1$ for $x=2$. Replacing $C$ by its value, transposing, gathering like terms and dividing by $x-2$, we get $-2x=A+B(x+1)$, which gives $A=-2$ for $x=-1$. Replacing $A$ by its value, we get

$2x=-2+B(x+1)$.

Hence $B=2$; so that the partial fractions are:

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

instead of ${\dfrac {1}{x+1}}+{\dfrac {x-1}{(x+1)^{2}}}+{\dfrac {1}{x-2}}$ stated above as being the fractions from which ${\dfrac {3x^{2}-2x+1}{(x+1)^{2}(x-2)}}$ was obtained. The mystery is cleared if we observe that ${\dfrac {x-1}{(x+1)^{2}}}$ can itself be split into the two fractions ${\dfrac {1}{x+1}}-{\dfrac {2}{(x+1)^{2}}}$, so that the three fractions given are really equivalent to

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

which are the partial fractions obtained.

We see that it is sufficient to allow for one numerical term in each numerator, and that we always get the ultimate partial fractions.

When there is a power of a factor of $x^{2}$ in the denominator, however, the corresponding numerators must be of the form $Ax+B$; for example,

${\frac {3x-1}{(2x^{2}-1)^{2}(x+1)}}={\frac {Ax+B}{(2x^{2}-1)^{2}}}+{\frac {Cx+D}{2x^{2}-1}}+{\frac {E}{x+1}}$,