Page:Calculus Made Easy.pdf/145

 If $$x=-3$$, we get:

So then the partial fractions are:

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:

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

hence

and

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

hence

and