Page:Calculus Made Easy.pdf/149

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OTHER USEFUL DODGES

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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,

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

C.M.E.

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