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126

Calculus Made Easy

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