Page:Calculus Made Easy.pdf/146

so that $A=1$, and the partial fractions are:

${\frac {x-1}{x^{2}+1}}-{\frac {2}{x+1}}$.

Take as another example the fraction

${\frac {x^{3}-2}{(x^{2}+1)(x^{2}+2)}}$.

We get

${\begin{aligned}{\frac {x^{3}-2}{(x^{2}+1)(x^{2}+2)}}&={\frac {Ax+B}{x^{2}+1}}+{\frac {Cx+D}{x^{2}+2}}\\&={\frac {(Ax+B)(x^{2}+2)+(Cx+D)(x^{2}+1)}{(x^{2}+1)(x^{2}+2)}}.\end{aligned}}$

In this case the determination of $A$, $B$, $C$, $D$ is not so easy. It will be simpler to proceed as follows: Since the given fraction and the fraction found by adding the partial fractions are equal, and have identical denominators, the numerators must also be identically the same. In such a case, and for such algebraical expressions as those with which we are dealing here, the coefficients of the same powers of $x$ are equal and of same sign.

Hence, since

${\begin{aligned}x^{3}-2&=(Ax+B)(x^{2}+2)+(Cx+D)(x^{2}+1)\\&=(A+C)x^{3}+(B+D)x^{2}+(2A+C)x+2B+D,\end{aligned}}$

we have $1=A+C$; $0=B+D$ (the coefficient of $x^{2}$ in the left expression being zero); $0=2A+C$; and $-2=2B+D$. Here are four equations, from which we readily obtain $A=-1$; $B=-2$; $C=2$; $D=0$; so that the partial fractions are ${\dfrac {2(x+1)}{x^{2}+2}}-{\dfrac {x+2}{x^{2}+1}}$.