Mathematics

Resolve $$\displaystyle \frac{x^2-4}{(x^2+1)(x^2+2)(x^2+3)}$$ into partial fractions.


ANSWER

$$\displaystyle -\frac{5}{2(x^2+1)}+\frac{6}{(x^2+2)}+\frac{7}{2(x^2+3)}$$


SOLUTION
The given fraction has everywhere $$x^2$$. Put $$x^2 = t$$ for the sake of partial fraction,
then $$\displaystyle \frac{x^2-4}{(x^2+1)(x^2+2)(x^2+3)}=\frac{t-4}{(t+1)(t+2)(t+3)}$$
Let $$\displaystyle \frac{t-4}{(t+1)(t+2)(t+3)}=\frac{A}{t+1}+\frac{B}{t+2}+\frac{C}{t+3}$$                         ...(i)
$$\therefore \displaystyle A=\lim_{t \rightarrow -1}(t+1)\left[\frac{(t-4)}{(t+1)(t+2)(t+3)}\right]$$
$$\displaystyle = \lim_{t \rightarrow -1}\left[\frac{(t-4)}{(t+2)(t+3)}\right]$$ $$\displaystyle = -\frac{5}{4}$$
$$\displaystyle B = \lim_{t \rightarrow -2}(t+2)\left[\frac{(t-4)}{(t+1)(t+2)(t+3)}\right]$$
$$\displaystyle =\lim_{t \rightarrow -2}\left[\frac{(t-4)}{(t+1)(t+3)}\right]$$ $$=6$$
and$$\displaystyle C= \lim_{t \rightarrow -3}(t+3)\left[\frac{(t-4)}{(t+1)(t+2)(t+3)}\right]$$
$$\displaystyle = \lim_{t \rightarrow -3}\left[\frac{(t-4)}{(t+1)(t+2)}\right] \displaystyle = \frac{-7}{(-2)(-1)}$$ $$\displaystyle = \frac{-7}{2}$$
Substituting the values of A, B and C in (i), we have
$$\displaystyle \frac{t-4}{(t-1)(t+2)(t+3)} = -\frac{5}{2(t+1)}+\frac{6}{(t+2)}+\frac{7}{2(t+3)}$$
In last put $$t=x^2$$
$$\therefore \displaystyle \frac{x^2-4}{(x^2+1)(x^2+2)(x^2+3)} = -\frac{5}{2(x^2+1)}+\frac{6}{(x^2+2)}+\frac{7}{2(x^2+3)}$$
Which are required partial fractions.
Hence, option 'C' is correct.
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