Partial fraction decomposition for $\int \frac{-x^2+x+4}{(1-x)^2(3x+1)} dx$

Question

We want to evaluate the integral

$$\int \frac{-x^2+x+4}{(1-x)^2(3x+1)} dx$$

What I have troubles with, is to understand the principle of partial fraction decomposition.

For instance, here we'd have

$$\frac{-x^2+x+4}{(1-x)^2 (3x+1)} = \frac{c_1}{(1-x)} + \frac{c_2}{(1-x)^2} + \frac{c_3}{(3x+1)}$$

Sure, we could say

$$ c_3 = \bigg(\frac{-x^2+x+4}{(1-x)^2 (3x+1)} - \frac{c_1}{(1-x)} - \frac{c_2}{(1-x)^2} \bigg) \cdot (3x+1)$$

but how does that help us?

How do we find out that

$c_3 = 2$, $c_2 = 1$ and $c_1 = 1$?

One of the denominator should be $(1-x)^2$ instead of $1-x$. — Arctic Char, Jul 27 '22 at 16:28
@ArcticChar No. The answer of that thread is difficult to understand and too long for a topic which should be coverable in a few minutes — stepbysteptomathpro, Jul 27 '22 at 16:32
Say, for a simpler expression like $\frac{2x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$, do you know how to find $A, B$? — Arctic Char, Jul 27 '22 at 16:43
@ArcticChar Yes. See https://math.stackexchange.com/questions/4501221/question-regarding-partial-fraction-decomposition-for-int-frac1x2-4-dx — stepbysteptomathpro, Jul 27 '22 at 16:43
Multiply the RHS,
$$c_3 = \frac{-x^2+x+4}{(1-x)^2} - \frac{c_1(3x+1)}{(1-x)} - \frac{c_2(3x+1)}{(1-x)^2}$$

Then substitute $x=-\frac13$ to find $c_3$. Similarly for $c_2$. — peterwhy, Jul 27 '22 at 16:46

Vasili · Answer 1 · 2022-07-27T17:16:38.267

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Start with: $$\frac{-x^2+x+4}{(1-x)^2 (3x+1)} = \frac{c_1}{(1-x)} + \frac{c_2}{(1-x)^2} + \frac{c_3}{(3x+1)}=$$$$\frac{c_1(1-x)(3x+1)+c_2(3x+1)+c_3(1-x)^2}{(1-x)^2 (3x+1)}=$$

$$\frac{c_1(-3x^2+2x+1)+c_2(3x+1)+c_3(1-2x+x^2)}{(1-x)^2 (3x+1)}=$$$$\frac{x^2(-3c_1+c_3)+x(2c_1+3c_2-2c_3)+c_1+c_2+c_3}{(1-x)^2 (3x+1)}$$ Next, use that two polynomials are equal when all coeddicients are equal so: $$ \left\{ \begin{array}{c} -3c_1+c_3=-1 \\ 2c_1+3c_2-2c_3=1 \\ c_1+c_2+c_3=4 \end{array} \right. $$ Solve the system and find what you need

edited Jul 27 '22 at 17:16

answered Jul 27 '22 at 16:47

Vasili

10,690

Thank you. What do you mean with this statement: "Polynomials are equal when all coefficients are equal"? How do you get to $-1$ looking at $-3c_1 + c_2$? – stepbysteptomathpro Jul 27 '22 at 16:52
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They have equal coefficients of corresponding powers of the independent variable. On the left side, coefficient for $x^2$ is $-1$, on the right side it's $-3c_1+c_3$ (had a typo) – Vasili Jul 27 '22 at 17:16
Sorry, I don't understand what you mean with "coefficient for $x^2$ is $-1$". Can you elaborate? – stepbysteptomathpro Jul 27 '22 at 17:24
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@stepbysteptomathpro If you do not understand the answer, why have you accepted it? – Arturo Magidin Jul 27 '22 at 17:25
@ArturoMagidin Because he put so much effort into it (and I'm just a retard for not understanding) – stepbysteptomathpro Jul 27 '22 at 17:27
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@stepbysteptomathpro: On the lhs we have $-x^2=(-1)\cdot x^2$ – Vasili Jul 27 '22 at 17:28
@Vasili But on the lhs for $c_1+c_2+c_3$ we have nothing. How do we get to $4$ there then? – stepbysteptomathpro Jul 27 '22 at 17:29
@stepbysteptomathpro Accepting an answer is a signal to the community that you are satisfied with the given answer and that your doubts have been completely addressed. It discourages other people from providing alternative answers or from expanding. Do not accept an answer until you are done. If you want to reward "effort", then up-vote. – Arturo Magidin Jul 27 '22 at 17:30
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@stepbysteptomathpro: No, we have $4 \cdot x^0$ on the left. It's all about powers of $x$ – Vasili Jul 27 '22 at 17:39
Not a very efficient method IMHO – David Quinn Jul 27 '22 at 18:04
@stepbysteptomathpro Please do not use the word retard. – Sean Roberson Jul 27 '22 at 18:08

score 1 · Answer 2 · answered Jul 27 '22 at 17:25

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Evaluate the coefficients directly \begin{align} c_2 = &\lim_{x\to 1}\frac{-x^2+x+4}{3x+1}=1\\ c_3 = &\lim_{x\to -\frac13}\frac{-x^2+x+4}{(1-x)^2}=2\\ c_1 = &\lim_{x\to \infty}\frac{-x^2+x+4}{(1-x)(3x+1)}-\frac{c_2}{1-x} -\frac{c_3(1-x)}{3x+1}=\frac13+\frac{c_3}3 =1\\ \end{align}

answered Jul 27 '22 at 17:25

Quanto

97,352

How do you know which limits to take? – stepbysteptomathpro Jul 27 '22 at 17:30

score 1 · Accepted Answer · answered Jul 27 '22 at 18:03

A simple and efficient way of doing this is as follows:

When you have $$\frac{-x^2+x+4}{(1-x)^2(3x+1)}=\frac{c_1}{1-x}+\frac{c_2}{(1-x)^2}+\frac{c_3}{3x+1}$$ Start by combining the RHS into one fraction with the same denominator as the LHS, and then just write out the numerators, i.e. $$-x^2+x+4\equiv c_1(1-x)(3x+1)+c_2(3x+1)+c_3(1-x)^2$$

Note that the $\equiv$ sign indicates that this is an identity and must be true for any value of $x$. Therefore choose values of $x$ which make brackets zero (to avoid a simultaneous equation).

When $x=1$, we have $4=0+c_2(4)+0\implies c_2=1$

When $x=-\frac13$, we have $\frac{32}{9}=0+0+c_3(\frac43)^2\implies c_3=2$

Now that we have run out of convenient values of $x$, we can just choose any other value, such as zero. Or, more conveniently, compare coefficients such as the $x^2$ term, which gives

$-1=-3c_1+c_3\implies c_1=1$

I hope this helps.

Oh that's a lot easier. Thank you! – stepbysteptomathpro Jul 27 '22 at 18:07 — stepbysteptomathpro, Jul 27 '22 at 18:07

score 0 · Answer 4 · answered Jul 28 '22 at 00:31

We first decompose the integrand into 2 pieces by letting $$\frac{-x^{2}+x+4}{(1-x)^{2} (3 x+1)} \equiv \quad \frac{f(x)}{(1-x)^{2}}+\frac{c_{3}}{3 x+1} \tag{*} $$ Then equating the numerators on both sides yields $$ -x^{2}+x+4 \equiv f(x)(3 x+1)+ c_{3}(1-x)^{2} \tag {**} $$ Putting $x=-\frac{1}{3}$ gives $\frac{32}{9}=\frac{16}{9} C_{3} \Rightarrow C_{3}=2$. Rearranging (**) yields $$ f(x)=\frac{-x^{2}+x+4-2(1-x)^{2}}{3 x+1}=-x+2=1-x+1 $$ Putting back to $(*)$ yields the decomposition as $$ \frac{-x^{2}+x+4}{\left(1-x^{2}\right)(3 x+1)} \equiv \quad \frac{1}{1-x}+\frac{1}{(1-x)^{2}}+\frac{2}{3 x+1} $$

Partial fraction decomposition for $\int \frac{-x^2+x+4}{(1-x)^2(3x+1)} dx$

4 Answers4