Question

partial fractions...

Answer 1

\[(x ^{2}+1)/x(x-1)^{3}\]

Answer 2

very weak algebra background..i see what it's suppose to look like but i don't get why so explain step by step pls?

Answer 3

\[=A/(x-1)+B/(x-1)^{2}+C/(x-1)^{3}+D/x\]
i understand how this is set up

Answer 4

i don't get why after you multiply the fractions by the denominator, it's\[Ax(x-1)^{2}+Bx(x-1)+Cx+D(x-1)^{3}\]

Answer 5

cuz i'm thinking you have to multiply each constant with each denominator. if not, how do you determine which denominator is multiplied by the constant??

Answer 6

You have the basic idea quite well
First we break it up as follows\[\frac{x^2+1}{x(x-1)^3}=\frac Ax+\frac B{x-1}+\frac C{(x-1)^2}+\frac D{(x-1)^3}\]now we multiply both sides by the denominator on the left, and set what's left over equal to the numerator\[A(x-1)^3+Bx(x-1)^2+Cx(x-1)+Dx=x^2+1\]now we have to distribute the left and collect like terms. We can then set the coefficients on the left and right equal and we get a system of equations\[A(x^3-3x^2+3x-1)+B(x^3-2x^2+x)+C(x^2-x)+Dx\]\[=(A+B)x^3+(-3A-2B+C)x^2+(3A+B-C+D)x-A\]\[=x^2+1\]Now pay attention the the coefficients on each side.
We get right away that\[-A=1\to A=-1\]which we can use to find B\[A+B=-1+B=0\to B=1\]which we can use to find C\[-3A-2B+C=3-2+C=1\to C=0\]Which we can use to find D\[3A+B-C+D=-3+1+D=0\to D=2\]and so we wind up with\[\frac{x^2+1}{x(x-1)^3}=\frac{-1}x+\frac1{x-1}+\frac2{(x-1)^3}\]I hope no typos!

Answer 7

can you answer my last question?

Answer 8

I'm not sure what you mean by multiply each denominator with the constant. The whole thing gets multiplied by \[x(x-1)^3\]
then we collect like terms, then compare coefficients on each side...
you don't always have to collect like terms, but in this case yes.

Answer 9

on the right hand side, add them together. You need a common denominator. Try it, and you see that the procedure is doing just that.

Answer 10

once you have everything on the right hand side over a common denominator, you equate numerators.

Answer 11

o you mult with the denominator from the left...i misunderstood that whole step, thank you