Mathematics

# Resolve $\displaystyle \frac{5x^3+6x^2+5x}{(x^2-1)(x+1)^3}$ into partial fractions.

$\displaystyle \frac{2}{(x+1)^4}-\frac{3}{(x+1)^3}+\frac{3}{(x+1)^2}-\frac{1}{(x+1)}+\frac{1}{x-1}$

##### SOLUTION
We have
$\displaystyle \frac{5x^3+6x^2+5x}{(x^2-1)(x+1)^3}=\frac{5x^3+6x^2+5x}{(x-1)(x+1)^4}$                                                 ...(i)
Putting $x+1=y$ i.e., $x=y-1$ in (i), we get
$\displaystyle \frac{5x^3+6x^2+5x}{(x-1)(x+1)^4}= \frac{5(y-1)^3+6(y-1)^2+5(y-1)}{(y-2)y^4}$

$\displaystyle= \frac{5(y^3-1-3y^2+3y)+6(y^2-2y+1)+5(y-1)}{(y-2)y^4}$

$\displaystyle= \frac{5y^3-9y^2+8y-4}{(y-2)y^4}$

$\displaystyle=\frac{(-4+8y-9y^2+5y^3)}{y^2(-2+y)}$                                            ...(ii)
Divide $-4+8y-9y^2+5y^3$ by $-2+y$ till the remainder is $y^4$.
$-2+y \downharpoonleft -4 +8y-9y^2+5y^3 \downharpoonright 2-3y+3y^2-y^3$
$-4 +2y$
$+ -$
_____________________
$6y-9y^2$
$6y-3y^2$
$- +$
_____________________
$-6y^2+5y^3$
$-6y^2+3y^3$
$+ -$
______________________
$2y^3$
$2y^3-y^4$
$- +$
_______________________
$y^4$
$\therefore \displaystyle \frac{-4+8y-9y^2+5y^3}{y^4(-2+y)}=\frac{1}{y^4}\left[2-3y+3y^2-y^3+\frac{y^4}{-2+y}\right]$

$\displaystyle= \frac{2}{y^4}-\frac{3}{y^3}+\frac{3}{y^2}-\frac{1}{y}+\frac{1}{-2+y}$
Putting $y=x+1$
Hence                  $\displaystyle \frac{5x^3+6x^2+5x}{(x-1)(x+1)^4}=\frac{2}{(x+1)^4}-\frac{3}{(x+1)^3}+\frac{3}{(x+1)^2}-\frac{1}{(x+1)}+\frac{1}{x-1}$
which are the 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 86

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