Mathematics

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


ANSWER

$$\displaystyle \frac{1}{2(x-1)^4}+\frac{7}{4(x-1)^3}+\frac{17}{8(x-1)^2}+\frac{15}{16(x-1)}+\frac{1}{16}\cdot\frac{1}{(x+1)}$$


SOLUTION
Let $$x-1=y$$, i.e., $$x=1+y$$
$$\therefore         \displaystyle \frac{x^4}{(x-1)^4(x+1)}= \frac{(1+y)^4}{y^4(1+y+1)}$$

                                        $$= \displaystyle \frac{1+4y+6y^2+4y^3+y^4}{y^4(2+y)}$$
Divide $$1+4y+6y^2+4y^3+y^4$$ by $$2+y$$ and continue till the remaider is $$y^4$$.
            $$2+y            1+4y+6y^2+4y^3+y^4 \downharpoonright  \frac{1}{2}+\frac{7}{4}y+\frac{17}{8}y^2+\frac{15}{16}y^3$$
                            $$1+\frac{1}{2}y$$
                         $$-        -$$
                        __________________
                                   $$\frac{7}{2}y+6y^2$$
                                   $$\frac{7}{2}y+\frac{7}{6}y^2$$
                                 $$-       -$$
                               _____________________
                                             $$\frac{17}{4}y^2+4y^3$$
                                             $$\frac{17}{4}y^2+\frac{17}{8}y^3$$
                                            $$-                   -$$
                                         ________________________
                                                            $$\frac{15}{8}y^3+y^4$$
                                                            $$\frac{15}{8}y^3+\frac{15}{16}y^4$$
                                                          $$-         -$$
                                                      ___________________
                                                                            $$\frac{1}{16}y^4$$
$$\therefore    \displaystyle \frac{1+4y+6y^2+4y^3+y^4}{(2+y)}=\left[\frac{1}{2}+\frac{7}{4}y+\frac{17}{8}y^2+\frac{15}{16}y^3+\frac{\frac{1}{16}y^4}{2+y}\right]$$

$$\therefore    \displaystyle \frac{(1+4y+6y^2+4y^3+y^4}{y^4(2+y)}=\frac{1}{2y^4}+\frac{7}{4y^3}+\frac{17}{8y^2}+\frac{15}{16y}+\frac{1}{16}\cdot\frac{1}{2+y}$$

in last putting $$y=x-1$$
$$\therefore    \displaystyle \frac{x^4}{(x-1)^4(x+1)} = \frac{1}{2(x-1)^4}+\frac{7}{4(x-1)^3}+\frac{17}{8(x-1)^2}+\frac{15}{16(x-1)}+\frac{1}{16}\cdot \frac{1}{(x+1)}$$
which are the required partial fractions.
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Single Correct Medium Published on 17th 09, 2020
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