Mathematics

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

$\displaystyle \frac{-1}{4(1+x)}+\frac{(x-1)}{4(1+x^2)}+\frac{(x+1)}{2(1+x^2)^2}$

##### SOLUTION
Let             $\displaystyle \frac{x}{(1+x)(1+x^2)^2}=\frac{A}{1+x}+\frac{Bx+C}{(1+x^2)}+\frac{(Dx+E)}{(1+x^2)^2}$                      ...(i)
$\Longrightarrow x=A(1+x^2)^2+(Bx+C)(1+x)(1+x^2)+(Dx+E)(1+x)$                ...(ii)
Putting $1+x=0$ or $x=-1$ in (ii), we obtain
$-1=A[1+(-1)^2]^2+0+0$
$\therefore A=-\dfrac{1}{4}$
Putting $1+x^2=0$ or $x^2= -1$ in (ii), we obtain
$x=0+0+Dx+D(-1)+E+Ex$
$\Longrightarrow x=(D+E)x+(E-D)$
Equating the coefficient of $x$ and constant term, we get
$D+E=1$
and    $E-D=0$
$\therefore D=E=\dfrac{1}{2}$
Comparing the constant terms in (ii), we obtain
$0=A+C+E$   (For comparing constant terms putting $x=0$)
or                 $0=-\dfrac{1}{4}+C+\dfrac{1}{2}$
$\Longrightarrow 0=C+\dfrac{1}{4}$
$\therefore C=-\dfrac{1}{4}$
and comparing the coefficient of $x^4$ in (ii), we obtain
$0=A+B$
$\therefore B= -A$
$B=\dfrac{1}{4}$
Substituting the value of A, B, C, D, and E in (1), then
$\displaystyle \frac{x}{(1+x)(1+x^2)^2}= \frac{-1}{4(1+x)}+\frac{(x-1)}{4(1+x^2)}+\frac{(x+1)}{2(1+x^2)^2}$
which are the required partial fractions.
Hence, option 'A' is correct.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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