Finding a faster method of partial fraction decomposition.

Question

While integrating: $$ \int \cot^{-1} (x^2-x+1) dx $$ I had to do a large number of partial fraction decompositions, the following functions: $$ \frac {x(1-2x)}{(x^2-x+1)^2+1}, $$ $$ \frac {x}{(x^2+1)(x^2-2x+2)} $$ $$ \frac {x^2}{(x^2+1)(x^2-2x+2)} $$ A similarly wild example is that of the integral of. $\sqrt \tan x$. Of course there may be a number of promising alternate methods of integration, but for me, personally, the method of partial fraction decomposition (p.f.d.) is a novel one, for it can simplify monstrous functions. The only problem with it is it's comparative slowness - yes, it is a TIME EATER. What I am seeking must be clear now, a faster technique of p.f.d. Apart from mainstream algorithms, I am also interested in what is known as 'Vedic Maths', which from my experience I shall call heuristic approaches. Moreover, I would love to learn how does a CAS do it within seconds. Any help would be greatly appreciated .

Answer 1

You can use Heaviside Cover-up method. The following example illustrates it:

$$\dfrac{3x^2+12x+11}{(x+1)(x+2)(x+3)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}+\dfrac{C}{x+3}$$

1. Finding $A$. Multiply both sides by $x+1$ and set $x=-1$. $$\dfrac{3x^2+12x+11}{(x+2)(x+3)}=A+\underbrace{\dfrac{B(x+1)}{x+2}+\dfrac{C(x+1)}{x+3}}_{=0 \text{ when }x=-1}$$
2. Finding $B$. Multiply both sides by $x+2$ and set $x=-2$. $$\dfrac{3x^2+12x+11}{(x+1)(x+3)}=B+\underbrace{\dfrac{A(x+2)}{x+1}+\dfrac{C(x+2)}{x+3}}_{=0 \text{ when } x=-2}$$
3. Finding $C$. Multiply both sides by $x+3$ and set $x=-3$. $$\dfrac{3x^2+12x+11}{(x+1)(x+2)}=C+\underbrace{\dfrac{A(x+3)}{x+1}+\dfrac{B(x+3)}{x+2}}_{=0 \text{ when }x=-3}$$

---

More Generally: In general, if the denominator of the proper rational function factors into the product of distinct linear factors:$$\dfrac{p(x)}{(x-a_1)(x-a_2)\cdots(x-a_r)}=\dfrac{A_1}{x-a_1}+\dfrac{A_2}{x-a_2}+\cdots+\dfrac{A_r}{x-a_r}$$then $A_i$'s found by covering up the factor $x-a_i$ the left, and setting $x=a_i$ the rest of the expression.

Answer 2

We can rigorous calculation as follows:

Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

and $$\arctan\dfrac{x-(x-1)}{1+x(x-1)}=?$$