Any shortcuts for integrating $\frac{x^6}{(x-2)^2(1-x)^5}$ by partial fractions?

Question

Is there a faster way to get the partial fraction decomposition of this $\frac{x^6}{(x-2)^2(1-x)^5}$?

$\frac{x^6}{(x-2)^2(1-x)^5} = \frac{A_1}{x-2} + \frac{A_2}{(x-2)^2} + \frac{B_1}{1-x} + \frac{B_2}{(1-x)^2} + \frac{B_3}{(1-x)^3} + \frac{B_4}{(1-x)^4} + \frac{B_5}{(1-x)^5}$

It's fairly easy to get $A_2$ and $B_5$, just by "covering up" $x-2$ and substituting $x=2$ and "covering up" $1-x$ and substituting $x=1$. So $A_2 = 64$ and $B_5=1$. But to proceed from there is there no other faster way other than to mulitply both sides of the equation by $(x-2)^2(1-x)^5$ and compare coefficients, which would result in a system of 6 equations with 6 unknowns?

Answer 1

The "cover-up" rule gives correct answers but it suffers greatly from lack of rigour, in fact no rigour at all. Here's a more mathematically proper way to find the coefficients.First multiply through by $$(x-2)^2(1-x)^5$$ Your new equation is $$x^6=A_1(x-2)(1-x)^5+A_2(1-x)^5+B_1(1-x)^4(x-2)^2+B_2(1-x)^3(x-2)^2+B_3(1-x)^2(x-2)^2+B_4(1-x)(x-2)^2+B_5(x-2)^2$$ Note that the original equation is true for all $x$ except 2 or 1 iff your new equatiion is true for all $x$ except possibly 2 or 1 iff your new equation is true for all x.[Two polynomials in one variable are equal for all $x$ iff they are equal for infinitely many $x$.There are infinitely many $x$ other than 1 or 2.]In this new equation put $x=2$ to find $A_1$ and then $x=1$ to find $B_5$Now take the derivative. $$6x^5=A_1(1-x)^5-5(1-x)^4(A_1(x-2)+A_2)+(-4B_1(1-x)^3-3B_2(1-x)^2-2B_3(1-x)-B_4)(x-2)^2+2(B_1(1-x)^4+B_2(1-x)^3+B_3(1-x)^2+B_4(1-x)+B_5)(x-2)$$ Put $x=2$ in the derived equation. Since you already know $A_2$ you can find $A_1$ Put $x=1$ in the derived equation. Since you already know $B_5$ you can find $B_4$. Another derivative will give you $B_3$ and so on.

Answer 2

set $$x=1-\frac{1}{u+1}$$ to get $$\int{\frac{{{x}^{6}}}{{{(x-2)}^{2}}{{(1-x)}^{5}}}dx}=\int{\frac{{{u}^{6}}}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}du}$$ long division for the integrand gives: $${{u}^{3}}-5{{u}^{2}}+17u-49+\frac{129{{u}^{2}}+324u+196}{\left( u+1 \right){{\left( u+2 \right)}^{2}}}$$

Answer 3

There exists a shortcut of a generic nature for this type of rational function with two linear factors in the denominator. For this particular case, I verified that the method does not contain any hidden complexities by doing it entirely by hand, so no computer algebra system used, not even a calculator.

The cover-up method is basically the Laurent expansion method. So, a rational function equals the sum of all the singular parts of the expansions around all singular points. This is also true if the degree of the numerator is not smaller than that of the denominator. In that case, the point at infinity is also a singular point, the singular part of the expansion around that point is the part of the expansion in positive powers of $x$, so that's then exactly the result of long division.

So, all we need to do is to expand around the poles at $x = 2$ and $x = 1$, and add up the truncated expansions that only contain terms with negative powers of $(x-2)$ and $x-1$ and we're done. But it's possible to simplify the computations by observing that if the degree of the numerator had been $1$ or $0$, that the expansion around infinity of the function would have started with a power of $x^{-1}$ that would have been larger than the largest power of the $(x-1)^{-1}$ term. If we had $x$ in the numerator, then the expansion around infinity starts with $x^{-6}$, while the part of the partial fraction expansion around $x-1$ always contains a largest power of $(x-1)^{-5}$.

This means that the part of the part of the partial fraction expansion around $x = 1$ can be obtained by expanding the partial fraction expansion around $x = 2$ around infinity, by expanding this in negative powers of $x-1$, which is much simpler to do that computing the expansion of the rational function around $x-1$ because you then need to expand the ratio of two terms.

In the following I'll take out a minus sign from $1-x$ in the denominator, so there will be an overall minus sign compared to the rational function as written down by the OP. Let's then start with writing down the partial fraction expansion of the function:

$$R(x) = \frac{x}{(x-2)^2(x-1)^5}$$

Let's then put $x = t+2$ and expand in powers of $t$ to obtain the singular part of the expansion around $x = 2$. This yields:

$$R(2+t) = \frac{2+t}{t^2 (1+t)^5} = \frac{1}{t^2}(2+t)\left[1-5 t +\mathcal{O}\left(t^2\right)\right] = \frac{2}{t^2}-\frac{9}{t}+ \mathcal{O}\left(1\right)$$

So, the part of the partial fraction expansion from the singularity at $x = 2$ is:

$$R_1(x) = \frac{2}{(x-2)^2}-\frac{9}{x-2}$$

As argued above, the part of the partial fraction expansion from the singularity at $x =1$, which we'll denote by $R_2(x)$, is entirely determined by $R_1(x)$, because $R(x)$ decays faster when $x$ is sent to infinity than all the the term in $R_2(x)$. So, $R_2(x)$ is minus the expansion of $R_1(x)$ in powers of $(x-1)^{-1}$ to fifth order. If we put $x = 1+t$ and expand around $t=\infty$, we get:

$$\begin{split}R_1(1+t) &= \frac{2}{(t-1)^2}-\frac{9}{t-1} = \frac{2}{t^2}\sum_{k=0}^3\frac{k+1}{t^k} -\frac{9}{t}\sum_{k=0}^4\frac{1}{t^k} \mathcal{O}\left(t^{-6}\right)\\ &= -\left[\frac{9}{t} + \frac{7}{t^2} + \frac{5}{t^3}+ \frac{3}{t^4}+\frac{1}{t^5}\right]+\mathcal{O}\left(t^{-6}\right)\end{split}$$

We thus have:

$$R_2(x) = \frac{9}{(x-1)} + \frac{7}{(x-1)^2} + \frac{5}{(x-1)^3}+ \frac{3}{(x-1)^4}+\frac{1}{(x-1)^5}$$

And we have $R(x) = R_1(x) + R_2(x)$. But we're not interested in $R(x)$, we want to expand $S(x) = x^5 R(x)$ in partial fractions. This is then given by the sum of the singular parts of the Laurent expansion of $S(x)$ around $x = 2$ and $x = 1$, which we'll denote by $S_1(x)$ and $S_2(x)$, respectively. To compute $S_1(x)$, we then need to expand $x^5 R_1(x)$ around $x = 2$. If we put $x = t+2$, we get:

$$S(2+t) = (2+t)^5 R_1(2+t)+\mathcal{O}(1)=(2+t)^5\left(\frac{2}{t^2}-\frac{9}{t}\right) +\mathcal{O}(1) = \frac{64}{t^2} -\frac{128}{t} +\mathcal{O}(1)$$

We thus have:

$$S_1(x) = \frac{64}{(x-2)^2} -\frac{128}{x-2}$$

To compute $S_2(x)$, we expand $x^5 R_2(x)$ around $x = 1$. If we put $x = t+1$, we get:

$$\begin{split}S(1+t) &= (1+t)^5 R_1(1+t) +\mathcal{O}(1)=(1+t)^5\left(\frac{1}{t^5}+\frac{3}{t^4} +\frac{5}{t^3} + \frac{7}{t^2} + \frac{9}{t}\right)+\mathcal{O}(1)\\ & = \frac{1}{t^5}+\frac{8}{t^4} +\frac{30}{t^3} + \frac{72}{t^2} + \frac{129}{t}+\mathcal{O}(1)\end{split}$$

We thus have:

$$S_2(x) = \frac{1}{(x-1)^5}+\frac{8}{(x-1)^4} +\frac{30}{(x-1)^3} + \frac{72}{(x-1)^2} + \frac{129}{x-1}$$

The partial fraction expansion is thus given by:

$$\begin{split}\frac{x^6}{(x-2)^2 (x-1)^5} &= \frac{64}{(x-2)^2} -\frac{128}{x-2} + \frac{1}{(x-1)^5}+\frac{8}{(x-1)^4} +\frac{30}{(x-1)^3} + \frac{72}{(x-1)^2}\\ &+ \frac{129}{x-1} \end{split}$$