Question

Resolve $\frac{(2x^2+3)(x^2+4)}{(3x^2+1)(5x^2+3)}$ into partial fractions.

• A. $\frac{2}{15}+\frac{77}{12(3x^2+1)}-\frac{153}{10(5x^2+3)}$
• B. $\frac{2}{15}+\frac{77}{12(3x^2+1)}-\frac{153}{20(5x^2+3)}$
• C. $\frac{2}{15}-\frac{77}{12(3x^2+1)}+\frac{153}{20(5x^2+3)}$
• D. $\frac{2}{15}-\frac{77}{12(3x^2+1)}+\frac{153}{10(5x^2+3)}$

Answer 1

The correct option is B $\frac{2}{15}+\frac{77}{12(3x^2+1)}-\frac{153}{20(5x^2+3)}$

In the given fraction everywhere $x^2$, put $x^2=t$ for the sake of partial fraction.
$\therefore \frac{(2x^2+3)(x^2+4)}{(3x^2+1)(5x^2+3)}=\frac{(2t+3)(t+4)}{(3t+1)(5t+3)}=\frac{2}{15}+\frac{f\left(t\right)}{(3t+1)(5t+3)}$
where $f\left(t\right)$ is the linear function of $t$
$\left(\because \frac{2.1}{3.5}=\frac{2}{15}\right)$
Let $\frac{(2t+3)(t+4)}{(3t+1)(5t+3)}=\frac{2}{15}+\frac{A}{3t+1}+\frac{B}{5t+3}$
...(i)
$\therefore A=\underset{x\to -\frac{1}{3}}{lim}(3t+1)\left[\frac{(2t+3)(t+4)}{(3t+1)(5t+3)}\right]$ $=\underset{t\to -\frac{1}{3}}{lim}\left[\frac{(2t+3)(t+4)}{(5t+3)}\right]$
$=\left[\frac{(-\frac{2}{3}+3)(-\frac{1}{3}+4)}{(-\frac{5}{3}+3)}\right]$ $=\frac{77}{12}$

and $B=\underset{t\to -\frac{3}{5}}{lim}(5t+3)\left[\frac{(2t+3)(t+4)}{(3t+1)(5t+3)}\right]$ $=\underset{t\to -\frac{3}{5}}{lim}\left[\frac{(2t+3)(t+4)}{(3t+1)}\right]$
$=\left[\frac{(-\frac{6}{5}+3)(-\frac{3}{5}+4)}{(-\frac{9}{5}+1)}\right]$ $=\frac{\frac{9}{5}\times \frac{17}{5}}{-\frac{4}{5}}$ $=-\frac{153}{20}$
Substituting the value of A and B in (i), then
$\frac{(2t+3)(t+4)}{(3t+1)(5t+3)}=\frac{2}{15}+\frac{77}{12}\cdot \frac{1}{(3t+1)}-\frac{153}{20}\frac{1}{(5t+3)}$
In last substituting $t=x^2$, we get
$\frac{(2x^2+3)(x^2+4)}{(3x^2+1)(5x^2+3)}=\frac{2}{15}+\frac{77}{12(3x^2+1)}-\frac{153}{20(5x^2+3)}$
which are required partial fractions.
Hence, option 'B' is correct.