Page:Calculus Made Easy.pdf/145

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OTHER USEFUL DODGES

125

If $x=-3$ , we get:

$16=(A\times {0})+(B\times {0})+C(-2\times {-4})$ ; whence $C=2$ .

So then the partial fractions are:

${\frac {3}{x+1}}-{\frac {1}{x-1}}+{\frac {2}{x+3}},$ ,

which is far easier to differentiate with respect to $x$ than the complicated expression from which it is derived.

Case II. If some of the factors of the denominator contain terms in $x^{2}$ , and are not conveniently put into factors, then the corresponding numerator may contain a term in $x$ , as well as a simple number; and hence it becomes necessary to represent this unknown numerator not by the symbol $A$ but by $Ax+B$ ; the rest of the calculation being made as before.

Try, for instance:

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

Putting $x=-1$ , we get $-4=C\times {2}$ ; and $C=-2$ ;

hence

$-x^{2}-3=(Ax+B)(x+1)-2x^{2}-2$ ;

and

$x^{2}-1=Ax(x+1)+B(x+1)$ .

Putting $x=0$ , we get $-1=B$ ;

hence

$x^{2}-1=Ax(x+1)-x-1$ or $x^{2}+x=Ax(x+1)$ ;

and

$x+1=A(x+1)$ ,

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