Partial fractions to integrate$\int \frac{4x^2 -20}{(2x+5)^3}dx$
Ruffini-Horner algorithm + Polynomial long division applied to partial fraction decomposition.
Because Ruffini-Horner is one of the most efficient methods to evaluate polynomials, this is probably one of the, if not the most efficient methods to compute this and any partial fraction decomposition in general. Cover up is a particular case of it, but very incomplete. That is why only learning cover up leaves out lots of exercises you will struggle with.
First let me show you the only computations you will need to do. They consist only in writing this table and this table alone.
. 4 0 -20
-5/2 . -10 25
.......................
. 4 -10 [5]
. -10
...........
. 4 [-20]
.....
. [4]
The numbers in brackets are the ones we were looking for.
CHECK my computations for mistakes afterwards. I often make computational mistakes.
Now I will explain where these numbers are coming from.
The method.
The method is based on Ruffini-Horner algorithm to evaluate polynomials. This is an algorithm to evaluate a polynomial at a given number. For example, evaluating $4x^2-20$ ar $x=-5/2$ efficiently. It also allows, combined with polynomial long division, for a calculation of any partial fraction decomposition in which the denominator splits completely. By this I mean that it factors completely and all its zeros are reals.
Assume that we have $\frac{P(x)}{Q(x)}$ (reduced fraction, i.e. $P$ and $Q$ have no common factors) and $(x-a)^n$ is the highest power of $(x-a)$ that divides $Q$. We want to write $$\frac{P(x)}{Q(x)}=\frac{A_1}{(x-a)}+\frac{A_2}{(x-a)^2}+\ldots+\frac{A_n}{(x-a)^n}+h(x),$$
where $h$ doesn't have a pole at $x=a$, i.e. it is the part of the partial fraction decomposition that doesn't have $(x-a)$ in the denominators. Let us multiply both sides by $(x-a)^n$. We get
$$(x-a)^n\frac{P(x)}{Q(x)}=A_1(x-a)^{n-1}+A_2(x-a)^{n-2}+\ldots+A_n+(x-a)^nh(x),$$
So, computing the coefficients $A_1,\ldots, A_n$ is the same as writing $\frac{(x-a)^nP(x)}{Q(x)}$ as a sum of powers of $(x-a)$.
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Writing a rational function as a sum of powers of $(x-a)$:
If we have a rational function $\frac{P(x)}{Q(x)}$ and the numerator and denominator are written in decreasing powers of $(x-a)$ then we can write is as a sum of powers of $(x-a)$ (very efficiently) by doing polynomial long division. You can read how this is done in the Wikipedia article. Just remember that here we need to organize the polynomials in decreasing powers of $(x-a)$ instead of ascending powers of $(x-a)$ as the usual long division does. The usual long division proceeds by dividing the terms of highest degree, but here we need to proceed by dividing the terms of lowest degree.
Therefore we only need to see how to (efficiently) write polynomials in powers of $(x-a)$ to complete the algorithm.
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(re)-Writing polynomials in powers of $(x-a)$:
This is the part where Ruffini-Horner is used.
Given a polynomial $P(x):=a_nx^n+\ldots+a_0$, say written in powers of $x$. We would like to compute
$$P(x)=A_0+A_1(x-a)+\ldots+A_n(x-a)^n.$$
Notice that $A_0$ is the result of computing $P(a)$. For this computation, one of the most efficient ways is to use Ruffini-Horner algorithm. The cool thing is that Ruffini doesn't only give you $P(a)$, but the partial computations you do give you the result of $(P(x)-A_0)/(x-a)$. It is clear that $$(P(x)-A_0)/(x-a)=A_1+A_2(x-a)+\ldots+A_n(x-a)^{n-1}.$$
Therefore, what we need to do is to apply again Ruffini to this new polynomial, who's coefficients we already have from the previous Ruffini. Repeating $n$ times you get all the $A_i$'s.
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Summarizing:
One of the most efficient ways to compute the partial fraction decomposition is to first do Ruffini enough times (the number of times corresponds to the degree of the polynomial) with the numerator and the denominator of the given fraction. This will give you the coefficients for writing the numerator and denominator as powers of $(x-a)$. Then do $n$ steps, where $n$ is the power to which $(x-a)$ appears in the denominator, of the long division with these two polynomials to get the coefficients you are looking for. DONE!
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Your Example:
I will show you using Ruffini. First learn how to do Ruffini and the little table used to compute. It is super easy and only involves adding and multiplying a few times. The link to wikipedia shows a short example from where you learn all there is to be learned about how to do Ruffini.
Well, first $(2x+5)^3$ can be written as $8(x+5/2)^3$ and we can forget about the $8$ because it goes outside the integral. So your integral essentially is
$$\int\frac{4x^2-20}{(x+5/2)^3}\text{d}x.$$
Now what we need is to write the numerator $4x^2-20$ in powers of $(x+5/2)$.
For this we use Ruffini. Write the coefficients of the numerator $4$, $0$ and $-20$ and do Ruffini with $-5/2$. Use a table as in the Wikipedia example.
In the Ruffinity table to get $4$, $-10$, $5$. This last number $5$ is the first coefficient of $4x^2-20$ written in powers of $(x+5/2)$, i.e.
$$4x^2-20=5+b(x+5/2)+c(x+5/2)^2.$$
Now, repeat Ruffini with the other numbers you got on the table before $5$, i.e. $4$ and $-10$, using the same evaluation at $-5/2$. We get $4$, and $-20$. This $20$ is the next coefficient, i.e. $b=-20$. Repeat Ruffini with the remaining number $4$, which is like doing nothing. You get $4$, which is the last number $c=4$.
This gives us $$4x^2-20=5-20(x+5/2)+4(x+5/2)^2.$$
Now we divide by the denominator $(x+5/2)^3$ to get $$\frac{4x^2-20}{(x+5/2)^3}=\frac{5}{(x+5/2)^3}+\frac{-20}{(x+5/2)^2}+\frac{4}{(x+5/2)}.$$
When you learn Ruffini you will see that all of this reduces to write down the little table with only $4$ multiplications and $3$ additions at the beginning of the post. I have friends that can do it in their heads.
. 4 0 -20
-5/2 . -10 25
.......................
. 4 -10 [5]
. -10
...........
. 4 [-20]
.....
. [4]
The numbers in brackets are the ones we were looking for.
.......................
Comparison with other methods commonly used.
'Cover up': Only works when the poles are of order $1$. In other words when the exponents of the factors of the denominator are one.
Indeterminate coefficients: These are the methods that suggest to multiply, expand and compare coefficients to get system of equations. These waste times in doing those algebraic manipulations, and afterwards you need to solve a system of equations. The system of equations is sometimes large, plus some students are not familiar with solving systems of equations.
- Other special tricks. These are good, but usually depend on the specific problem and are not applicable in general. For example, the 'cover up' method.
$$\int \frac{4x^2 -20}{(2x+5)^3}dx = \int \dfrac A{(2x + 5)} + \dfrac B{(2x + 5)^2} + \dfrac C{(2x + 5)^3}\, dx$$
Partial fractions can be thought of as the reverse of "finding the common denominator".
That is, we can equate the numerator of the original integrand with the expansion of each of the numerators of the desired fractions, times the factor we'd need to find the common denominator, if we were adding the fractions in the desired integrand.
Doing this gives us $$A(2x + 5)^2 + B(2x + 5) + C = 4x^2 - 20\tag{1}$$
Solve for C first: to do this we can put $x = -\frac 52$ to zero out all but $$C = 4(-5/2)^2 - 20 = 4\cdot \frac {25}{4} - 20 = 25 - 20 = 5$$
Then expand the left-hand side, substitute $C = 5$, and match up coefficients, (you can easily find $A$ that way, since it will be the only term on the left hand side that is a coefficient of a term with $x^2$. Substitute, solve for $B$ knowing $A$. In other situations, you could use two arbitrary values for $x$ to obtain two equations in two unknowns.
Expanding $(1)$: $$A(4x^2 + 20 x + 25) + B(2x + 5) + 5 = 4x^2 - 20 \tag{2}$$ $$\implies 4Ax^2 = 4 \implies A = 1$$
Now, we have $A = 1, C = 5$, and can substitute this into our equation $(2)$:
$$4Ax^2 + 20 Ax + 25A + 2Bx + 5B + 5 = 4x^2 - 20$$ $$ \iff (20 A + 2B)x = 0\cdot x $$ $$\iff 20 A + 2B = 0 $$ $$\iff 20\cdot 1 + 2B = 0 \iff B = -10 $$
$$\int \frac{4x^2 -20}{(2x+5)^3}dx = \int \dfrac 1{(2x + 5)} + \dfrac {-10}{(2x + 5)^2} + \dfrac 5{(2x + 5)^3}\, dx$$