Question

partial fraction

\(\large\tt \color{black}{\dfrac{x^5-1}{x^2(x^3+1)}}\)

Answer 1

\[\frac{x^5-1}{x^2 \left(x^3+1\right)}=\frac{2 x-1}{3 \left(x^2-x+1\right)}-\frac{1}{x^2}-\frac{2}{3 (x+1)}+1 \]

Answer 2

i dont get this
can u tell the method

Answer 3

@iambatman ,

Answer 4

first step is to make numerator's degree lesser than denominator's

Answer 5

\(\large\tt \color{black}{\dfrac{x^5-1}{x^2(x^3+1)}=1-\dfrac{x^2}{x^2(x^3+1)}}\)

Answer 6

thats a good start,
but nobody would leave x^2 like that without canceling

Answer 7

this one \(\large\tt \color{black}{\dfrac{x^5-1}{x^2(x^3+1)}=1-\dfrac{(x^2+1)}{x^2(x^3+1)}}\)

Answer 8

looks good

Answer 9

\[\large \dfrac{(x^2+1)}{x^2(x^3+1)} =\dfrac{(x^2+1)}{x^2(x+1)(x^2-x+1)} \]

Answer 10

just factored x^3+1 using a^3+b^3 formula ^^

Answer 11

see if that makes sense so far

Answer 12

yes i got till now

Answer 13

good, next refer this page
http://www.mathsisfun.com/algebra/partial-fractions.html

Answer 14

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

Answer 15

solve the constants A,B,C,D,E

Answer 16

in place of B should be Bx + constant here ?

Answer 17

because x^2 ?

Answer 18

Regarding the \(\dfrac{\cdots}{x^2}\) term... If you set the numerator to a linear factor \(bx+c\), you'll eventually find that \(b=0\), so it's the same as simply using the constant factor \(B\) as @ganeshie8 has it.

For future reference, given a factor with multiplicity > 1, you can use the same numerator as for lower order terms. Examples: (Let \(P(x)\) be a polynomial)
\[\frac{P(x)}{x^2}=\frac{A}{x}+\frac{B}{x^2}\]
\[\frac{P(x)}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}\]
\[\frac{P(x)}{x^2(x^2+1)^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+1}+\frac{Ex+F}{(x^2+1)^2}\]

Answer 19

that i was missing
thanks very much both