Question

Express it in partial fraction
15) 1/x^2(x^-1)
=
@ganeshie8

Answer 1

check the function again, looks there is some typo..

Answer 2

1/x^2(x^-1)

Answer 3

= A/x + B/x^2 + C/(x+1) + D/(x-1)
Do you know this method..? Can u teach me how to make all the denominators same?

Answer 4

that `^` is throwing me off :/

Answer 5

sorry

Answer 6

your function is like below :\[\large \dfrac{1}{x^2(x-1)}\]

?

Answer 7

jst realised it!
it's 1/x^2(x^2-1)

Answer 8

\[\large \dfrac{1}{x^2(x^2-1)}\]

Answer 9

yes..

Answer 10

\[\large \dfrac{1}{x^2(x^2-1)} = \dfrac{1}{x^2(x+1)(x-1)}\]

Answer 11

\[\large \dfrac{1}{x^2(x+1)(x-1)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x+1}+ \dfrac{D}{x-1}\]

Answer 12

so your setup looks good so far :)

Answer 13

your goal is to find those constants A,B,C,D

Answer 14

I want to clarify something...

Answer 15

If

Answer 16

sure, ask..

Answer 17

Thnks. If there's another question like 1/x^3(x+1)(x-1)...How wld the setup be

Answer 18

you can guess i hope :)

Answer 19

\[\dfrac{1}{x^3(x+1)(x-1)} = \dfrac{A}{x} + \dfrac{B}{x^2} +\dfrac{C}{x^3} + \dfrac{D}{x+1}+ \dfrac{E}{x-1} \]

Answer 20

adds up one more constant ^

Answer 21

Okay.

Answer 22

Let's continue wth the first question..

Answer 23

yeah familiar with `Cover up` method ?

Answer 24

I'm nt sure. You can show it to me..

Answer 25

its okay, lets work it using the method known to you :
comparing coefficients

Answer 26

\[\large \dfrac{1}{x^2(x+1)(x-1)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x+1}+ \dfrac{D}{x-1}\]

Answer 27

multiply the right hand side by a special kind of \(\large \color{Red}{1}\) :

\(\large \color{red}{ \dfrac{x^2(x+1)(x-1)}{x^2(x+1)(x-1)}}\)

Answer 28

after that

Answer 29

\[ \dfrac{1}{x^2(x+1)(x-1)} = \large \color{red}{ \dfrac{x^2(x+1)(x-1)}{x^2(x+1)(x-1)}} \left[\dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x+1}+ \dfrac{D}{x-1}\right]\]

Answer 30

\[ \dfrac{1}{x^2(x+1)(x-1)} = \large \color{red}{ \dfrac{1}{x^2(x+1)(x-1)}}\\ \left[A\color{red}{x(x+1)(x-1)} + \\ B\color{red}{(x+1)(x-1)} + \\ C\color{red}{x^2(x-1)} \\D\color{red}{x^2(x+1)} \right]\]

Answer 31

canceling denominator both sides gives :

\[1 = Ax(x+1)(x-1) +B(x+1)(x-1) + Cx^2(x-1) + Dx^2(x+1)\]

Answer 32

see if that looks okay so far

Answer 33

How to get this... Ax(x+1)(x-1)

Answer 34

after distributing, the first termis :

\(\large \color{Red}{x^2(x+1)(x-1)}\dfrac{A}{x}\)

Answer 35

right ?

Answer 36

yes ... So, I need to distribute each constant

Answer 37

yes, we do that to get rid off of the fractions

Answer 38

notice that the denominator x cancels out

Answer 39

\(\large \color{Red}{x^{\not 2}(x+1)(x-1)}\dfrac{A}{\not x}\)

Answer 40

leaving you with \( \large A \color{Red}{x(x+1)(x-1)} \)

Answer 41

what's nxt

Answer 42

compare the coefficients both sides

Answer 43

\[1 = Ax(x+1)(x-1) +B(x+1)(x-1) + Cx^2(x-1) + Dx^2(x+1)\]

Answer 44

or may be plugin some values

Answer 45

plugin x = 1, and see what you get

Answer 46

Constant A,B,C = 0

Answer 47

1=2D
D=1/2

Answer 48

What do I need to do after getting the values of all the constant

Answer 49

Yes !
next think of another good value to plugin

Answer 50

okay

Answer 51

after getting all the constants, you're done !

Answer 52

The goal of partial fractions is to split the given rational functions into separate fractions, thats all.

Answer 53

Understood!

Answer 54

\[\large \dfrac{1}{x^2(x+1)(x-1)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{x+1}+ \dfrac{D}{x-1}\]

Answer 55

Thnx @ganeshie8

Answer 56

Once you have the values of all the constants, just plug them above ^ and conclude !

Answer 57

when x=1
D=1/2
when x=-1
C=-1/2
How to find A and B

Answer 58

ganshie is currently offline that your problem

Answer 59

but just use the formula he gave u to plug in the numbers

Answer 60

im sorry but i am no that good at math but i am your go to person for history

Answer 61

By the way, the topic is about partial fraction

Answer 62

Okay, I'll wait for Ganesh coz I've nt learn matrix method to solve such questiions. But, u can show ur working here using matrices