Page:Calculus Made Easy.pdf/145

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

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