Solve Second Degree Polynomial for Rational Function

[haleycomet2]

Homework Statement

Find the second degree polynomial P(x)
such that P(0)=1,P'(0)=0,and
\intP(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 don't 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]

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.

 

[haleycomet2]

\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]

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]

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.

 

[Dick]

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, got to confess being lazy not to check the rest.

 

[haleycomet2]

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!