# Rational function

## Homework Statement

Find the second degree polynomial P(x)
such that P(0)=1,P'(0)=0,and
$\int$P(x)/{x3(x-1)2} dx

is a rational function

## Homework Equations

this chapter is about integration techniques,L'Hopital's Rule, and Improper Integral
partial fraction,partial integration are learnt.

## The Attempt at a Solution

Since P(x) is second degree polynomial,i let P(x)=Ax2+Bx+C
P(0)=1,so 1=A(0)+B(0)+c
C=1
P'(0)=0,so 2Ax+B=0,
B=0
So I know P(x) is Ax2+1,but i dont know how to do next on.Partial fraction is seem too many variables and have only 1 clue{P(0)=1},partial integration is hard too.I think "Rational function" is the key of this question,but i know rational fraction no more than it can be express as a/b,how to solve this question ?

Dick
Homework Helper
Partial fractions is the correct answer. Try and express the integrand in terms of partial fractions and find the value of A that will make the terms that would lead to answers in the integral that are not rational functions like logs, zero.

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$\frac{Ax^2 +1}{x^3(x-1)^2}$=$\frac{B}{x}$+$\frac{C}{x^2}$+$\frac{D}{x^3}$+$\frac{E}{x-1}$+$\frac{F}{(x-1)^2}$

It is clear that constant B and E have to be zero so that P(x) is rational.However.the value of A is different when solving the equation,after eliminating other variable and only one variable(B or E) is left.Is this my mistake??
ps;the answer is -3x2+1,while my answer is 3(using B) and 9(using E)....
Is there any software or website can solve such a question?It is torturous to do for such a long time and the final answer is wrong....... Below is my attempt:

Ax2+1=Bx2(x-1)2+Cx(x-1)2+D(x-1)2+Ex3(x-1)+Fx3

when x=1,Ax+1=F
x=0, 1=D
x=-1,
A+1=4B-4C+4D+2E-F
2A+2=4B-4C+4+2E (F=Ax+1,D=1)
A+1=2B-2C+2+E
A-1=2B-2C+E-----(1)

x=2,
1+4A=4B+2C+D+8E+8F
4A+1=4B+2C+1+8E+8A+8 (F=Ax+1,D=1)
-4A-8=4B+2C+8E
-2A-4=2B+C+4E----(2)

x=-2,
4A+1=4B(9)+(-2C)+9D-8E(-3)-8F
4A+1=36B-18C+9D+24E-8F
4A+1=36B-18C+9+24E-8A-8
-4A=36B-18C+24E
-2A=18B-9C+12E-----(3)

(2)*9==》-2A-4=2B+C+4E
(1)-(2)*9,
16A+36=18C-24E
8A+18=-9C-12E--(4）

(3）-（1）*9,
-11A+9=9C+3E-----(5)

(4)+(5),
-3A+27=-9e
-A+9=-3E
when A=9,E=0

Similarly,
(2)-(1)*4,
-2A=-2B+3C---(6)

(3)-(2)*6,
A+3=3B-3C---(7)

(6)+(7)
-A+3=B
when A =3, B=0

Dick
Homework Helper
No problem with what you are trying to do and how you are trying to do it, but the process is long enough that it's easy to make a mistake. I use a software package called 'maxima'. Type 'partfrac((A*x^2+1)/(x^3*(x-1)^2),x)'. You'll see you should have gotten the same condition for setting B and E to zero. Namely A+3=0 giving you the books answer.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Ax2+1=Bx2(x-1)2+Cx(x-1)2+D(x-1)2+Ex3(x-1)+Fx3

x=0 gives D=1 .

Multiply out the above expression, and equate coefficients for the various powers of x.

You can also use WolframAlpha to do the partial fraction expansion for x2/(x3(x-1)2) and 1/(x3(x-1)2) separately, then combine the results.

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Dick
Homework Helper
Ax2+1=Bx2(x-1)2+Cx(x-1)2+D(x-1)2+Ex3(x-1)+Fx3

x=0 gives D=1 .

Multiply out the above expression, and equate coefficients for the various powers of x.

You can also use WolframAlpha to do the partial fraction expansion for x2/(x3(x-1)2) and 1/(x3(x-1)2) separately, then combine the results.

haleycomet2 got if x=0, then D=1 in section (1). I didn't get past section (3), but everything looked ok up till then. Must be some mistake after, gotta confess being lazy not to check the rest.

No problem with what you are trying to do and how you are trying to do it, but the process is long enough that it's easy to make a mistake. I use a software package called 'maxima'. Type 'partfrac((A*x^2+1)/(x^3*(x-1)^2),x)'. You'll see you should have gotten the same condition for setting B and E to zero. Namely A+3=0 giving you the books answer.

What a brilliant software!!I can use this to check my work in future.Thank you!!:rofl: