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 $c_1 = 1$?

Answer 1

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.

Answer 2

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

Answer 3

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}

Answer 4

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 $$ \frac{-x^{2}+x+4}{\left(1-x^{2}\right)(3 x+1)} \equiv \quad \frac{1}{1-x}+\frac{1}{(1-x)^{2}}+\frac{2}{3 x+1} $$