Page:Calculus Made Easy.pdf/144

 But there is another way out of this difficulty. The equation must be true for all values of $$x$$; therefore it must be true for such values of $$x$$ as will cause $$x-1$$ and $$x+1$$ to become zero, that is for $$x=1$$ and for $$x=-1$$ respectively. If we make $$x=1$$, we get $$4=(A\times 0)+(B\times 2)$$, so that $$B=2$$; and if we make $$x=-1$$, we get $$-2=(A\times -2)+(B\times 0)$$, so that $$A=1$$. Replacing the $$A$$ and $$B$$ of the partial fractions by these new values, we find them to become $$\dfrac{1}{x+1}$$ and $$\dfrac{2}{x-1}$$; and the thing is done.

As a farther example, let us take the fraction $$\dfrac{4x^2+2x-14}{x^3+3x^2-x-3}$$. The denominator becomes zero when $$x$$ is given the value $$1$$; hence $$x-1$$ is a factor of it, and obviously then the other factor will be $$x^2+4x+3$$; and this can again be decomposed into $$(x+1)(x+3)$$. So we may write the fraction thus:

making three partial factors.

Proceeding as before, we find

Now, if we make $$x=1$$, we get:

If $$x=-1$$, we get: