Page:Calculus Made Easy.pdf/149

 which gives

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

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:

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

The partial fractions are, therefore,