Question

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

• A. $-\frac{3}{2(x^2+1)}+\frac{6}{(x^2+2)}+\frac{5}{2(x^2+3)}$
• B. $-\frac{5}{2(x^2+1)}+\frac{3}{(x^2+2)}-\frac{7}{2(x^2+3)}$
• C. $-\frac{5}{2(x^2+1)}+\frac{6}{(x^2+2)}+\frac{7}{2(x^2+3)}$
• D. $-\frac{5}{2(x^2+1)}+\frac{6}{(x^2+2)}+\frac{5}{2(x^2+3)}$

Answer 1

The correct option is C $-\frac{5}{2(x^2+1)}+\frac{6}{(x^2+2)}+\frac{7}{2(x^2+3)}$

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