MHB Expanding the Fraction 1/x(x+1)^2: A Quick Guide for Scientists

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The discussion centers on the expansion of the fraction 1/(x(x+1)^2) using partial fraction decomposition. Participants detail the process of determining coefficients A, B, and C by equating coefficients after multiplying through by the least common denominator. The final expanded form is confirmed as 1/x - 1/(x+1) - 1/(x+1)^2. There is also a conversation about the educational background regarding partial fractions, with some users noting that they were taught various cases for decomposition in precalculus. The thread emphasizes the importance of understanding the derivation of these methods rather than relying solely on calculators.
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$\frac{1}{x\left(x+1\right)^2}
=\frac{-1}{x+1}-\frac{1}{\left(x+1\right)^2}+\frac{1}{x}$

I tried for an hour but no
 
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Using partial fraction decomposition, we may write:

$$\frac{1}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$

Multiply through by $x(x+1)^2$ to get:

$$1=A(x+1)^2+Bx(x+1)+Cx=A(x^2+2x+1)+B(x^2+x)+Cx=(A+B)x^2+(2A+B+C)x+A$$

Equating coefficients, we obtain the system:

$$A+B=0$$

$$2A+B+C=0$$

$$A=1\implies B=-1\implies C=-1$$

Hence:

$$\frac{1}{x(x+1)^2}=\frac{1}{x}+\frac{-1}{x+1}+\frac{-1}{(x+1)^2}=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2}$$
 
Has anyone else seen this approach to Partial Fractions?

\frac{1}{x(x+1)^2}\;=\;\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}

Multiply through by the LCD: .1 \;=\;A(x+1)^2+Bx(x+1)+CxNow select values of x:

. . \text{Let }x = -1:\;1 \:=\:A(0) + B(0) + C(-1) \quad\Rightarrow\quad \boxed{C = -1}

. . \text{Let }x = 0:\;\;1 \:=\:A(1) + B(0) + C(0) \quad\Rightarrow\quad \boxed{A = 1}

. . \text{Let }x = 1:\;\;1 \:=\:A(4) +B(2) + C(1) \quad\Rightarrow\quad 1\:=\: (1)4 + 2B +(-1)1 \quad\Rightarrow\quad \boxed{B = -1}See? . Isn't this easier and faster?

 
Why is the demonator expansion of
$$x\left(x+1 \right)^2 $$
$$x, \ \ x+1, \ \ \left(x+1\right)^2 $$
 
When you were given this problem, did you understand what "expand this fraction" meant? That is, had you been given instruction is "partial fractions"?
 
this was never covered in class i just have textbook examples and forums
 
When I was a student in Precalculus/Calc II, we were given various forms for partial fraction decomposition without proof:

Suppose $P(x)$ is a polynomial whose degree is less than that of $Q(x)$.

Case I: Nonrepeated Linear Factors

$$\frac{P(x)}{Q(x)}=\frac{P(x)}{\prod\limits_{i=1}^{n}(a_ix+b_i)}=\sum_{i=1}^{n}\left(\frac{C_i}{a_ix+b_i}\right)$$

Case II: Repeated Linear Factors

$$\frac{P(x)}{Q(x)}=\frac{P(x)}{(ax+b)^n}=\sum_{i=1}^{n}\left(\frac{C_i}{(ax+b)^i}\right)$$

Combining the Cases: When the denominator $Q(x)$ contains distinct as well as repeated linear factors, then we combine the two above cases.

Case III: Nonrepeated Irreducible Quadratic Factors

$$\frac{P(x)}{Q(x)}=\frac{P(x)}{\prod\limits_{i=1}^{n}(a_ix^2+b_ix+c_i)}=\sum_{i=1}^{n}\left(\frac{C_ix+D_i}{a_ix^2+b_ix+c_i}\right)$$

Case IV: Repeated Irreducible Quadratic Factors

$$\frac{P(x)}{Q(x)}=\frac{P(x)}{(ax^2+bx+c)^n}=\sum_{i=1}^{n}\left(\frac{C_ix+D_i}{(ax^2+bx+c)^i}\right)$$

All of the above 4 cases may be combined.
 
you did this in precalculus?
 
karush said:
you did this in precalculus?

Yes, we just didn't apply it to integration then...we were told it would be used later on in calculus, but that it was in general a good skill to learn. :)
 
  • #10
karush said:
Why is the demonator expansion of
$$x\left(x+1 \right)^2 $$
$$x, \ \ x+1, \ \ \left(x+1\right)^2 $$
The reason that expansion works is that, for any A, B, C,
\frac{A}{x}+ \frac{B}{x+ 1}+ \frac{C}{(x+ 1)^2}= \frac{A(x+ 1)^2}{x(x+ 1)^2}+ \frac{Bx(x+1)}{x(x+ 1)^2}+ \frac{Cx}{x(x+ 1)^2}
= \frac{Ax^2+ 2Ax+ A}{x(x+1)^2}+ \frac{Bx^2+ Bx}{x(x+1)^2}+ \frac{Cx}{x(x+ 1)^2}
= \frac{(A+ B)x^2+ (2A+ B+ C)x+ A}{x(x+ 1)^2}
And the three equations, A+ B= p, 2A+ B+ C= q, A= r, are independent so can be solved for A, B, and C for any p, q, and r.

Notice what happens if we try just \frac{A}{x}+ \frac{B}{(x+ 1)^2}. We get \frac{Ax^2+ 2Ax+ A+ Bx}{x(x+ 1)^2}= \frac{Ax^2+ (2A+ B)x+ A}{x(x+ 1)^2} and the equations A= p, 2A+ B= q, A= r are not independent.
 
  • #11
thank you, that helped a lot
i normally just use the expand() on the TI not knowing how it was derived.😎
 
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