Question

Resolve $\frac{3 x^2+7}{x^4-3 x^2+2}$ into partial fractions.

• A. $\frac{10}{x^2-2}+\frac{5}{x-1}-\frac{5}{x+1}$
• B. $\frac{13}{x^2-2}+\frac{5}{x+1}-\frac{5}{x-1}$
• C. $\frac{5}{x^2-2}-\frac{10}{x-1}+\frac{10}{x+1}$
• D. $\frac{5}{x-1}-\frac{5}{x+1}-\frac{13}{x^2-2}$

Answer 1

Answer: (B)

Let $x^2=p$, then $\frac{3 x^2+7}{x^4-3 x^2+2}=\frac{3 p+7}{p^2-3 p+2}$
Let $\frac{3 p+7}{p^2-3 p+2}=\frac{A}{p-1}+\frac{B}{p-2}$
$\Rightarrow \frac{3 p+7}{(p-1)(p-2)}=\frac{A(p-2)+B(p-1)}{(p-1)(p-2)}$
Consider,
$3 p+7=A(p-2)+B(p-1) $
$\text { Put } p=1,-A=10 \Rightarrow A=-10 $
$p=2, B=13 $
$\frac{3 p+7}{p^2-3 p+2}=\frac{13}{p-2}-\frac{10}{p-1} .$
But $p=x^2$
$\therefore \frac{3 x^2+7}{x^4-3 p^2+2} =\frac{13}{x^2-2}-\frac{10}{x^2-1} . $
$ =\frac{13}{x^2-2}-5[\frac{1}{x-1}-\frac{1}{x+1}] $
$ =\frac{13}{x^2-2}+\frac{5}{x+1}-\frac{5}{x-1} .$