Mathematics

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

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

##### 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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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int \displaystyle \frac{xdx}{\sqrt{1+x^2+\sqrt{(1+x^2)^3}}}$ is equal to
• A. $\displaystyle \frac{1}{2}\ln \left ( 1+\sqrt{1+x^2} \right )+c$
• B. $2\left ( 1+\sqrt{1+x^2} \right )+c$
• C. None of these
• D. $2\sqrt{1+\sqrt{1+x^2}+c}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
The value of $\displaystyle\int \dfrac{x^2+5x-1}{\sqrt{x}}dx$ equals?

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate the following integral:
$\displaystyle\int^a_0\dfrac{x}{\sqrt{a^2+x^2}}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate:
$\displaystyle\int_{\tfrac{-3\pi}{2}}^{\tfrac{-\pi}{2}}{\left[{\left(x+\pi\right)}^{3}+{\cos}^{2}{\left(x+3\pi\right)}\right]dx}$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
The value of $\displaystyle\int\limits_{0}^{\dfrac{\pi}{2}}\dfrac{2^{\sin x}}{2^{\sin x}+2^{\cos x}}dx$ is
• A. $\pi$
• B. $0$
• C. none of these
• D. $\dfrac{\pi}{4}$