Mathematics

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


ANSWER

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


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
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