Page:Calculus Made Easy.pdf/144

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

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

making three partial factors.

Proceeding as before, we find

Now, if we make ${\displaystyle x=1}$, we get:

If ${\displaystyle x=-1}$, we get: