Partial fraction expansion of 1/x(x+1)² doesn't match my answer

[karush]

Expand
$\frac{1}{x\left(x+1\right)^2}
=\frac{-1}{x+1}-\frac{1}{\left(x+1\right)^2}+\frac{1}{x}$

I tried for an hour but no

 

[MarkFL]

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:

$$\frac{1}{x(x+1)^2}=\frac{1}{x}+\frac{-1}{x+1}+\frac{-1}{(x+1)^2}=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2}$$

 

[soroban]

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: .$$1 \;=\;A(x+1)^2+Bx(x+1)+Cx$$Now 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?

 

[karush]

Why is the demonator expansion of
$$x\left(x+1 \right)^2 $$
$$x, \ \ x+1, \ \ \left(x+1\right)^2 $$

 

[HallsofIvy]

When you were given this problem, did you understand what "expand this fraction" meant? That is, had you been given instruction is "partial fractions"?

 

[karush]

this was never covered in class i just have textbook examples and forums

 

[MarkFL]

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.

 

[karush]

you did this in precalculus?

 

[MarkFL]

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. :)

 

[HallsofIvy]

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.

 

[karush]

thank you, that helped a lot
i normally just use the expand() on the TI not knowing how it was derived.