Page:Calculus Made Easy.pdf/146

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

Take as another example the fraction

We get

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

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