Page:Calculus Made Easy.pdf/149

which gives

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

For $x=-1$, this gives $E=-4$. Replacing, transposing, collecting like terms, and dividing by $x+1$, we get

$16x^{3}-16x^{2}+3=2Cx^{3}+2Dx^{2}+x(A-C)+(B-D)$.

Hence $2C=16$ and $C=8$; $2D=-16$ and $D=-8$; $A-C=0$ or $A-8=0$ and $A=8$, and finally, $B-D=3$ or $B=-5$. So that we obtain as the partial fractions:

${\frac {(8x-5)}{(2x^{2}-1)^{2}}}+{\frac {8(x-1)}{2x^{2}-1}}-{\frac {4}{x+1}}$.

It is useful to check the results obtained. The simplest way is to replace $x$ by a single value, say $+1$, both in the given expression and in the partial fractions obtained.

Whenever the denominator contains but a power of a single factor, a very quick method is as follows:

Taking, for example, ${\dfrac {4x+1}{(x+1)^{3}}}$, let $x+1=z$; then $x=z-1$.

Replacing, we get

${\frac {4(z-1)+1}{z^{3}}}={\frac {4z-3}{z^{3}}}={\frac {4}{z^{2}}}-{\frac {3}{z^{3}}}$.

The partial fractions are, therefore,

${\frac {4}{(x+1)^{2}}}-{\frac {3}{(x+1)^{3}}}$.