How to quickly solve partial fractions equation? Often I am dealing with an integral of let's say:
$$\int\frac{dt}{(t-2)(t+3)}$$
or
$$\int \frac{dt}{t(t-4)}$$
or to make this a more general case in which I am interested the most:
$$\int \frac{dt}{(t+\alpha)(t+\beta)} \quad \quad \alpha, \beta \in \mathbb{R}$$
Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.
What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):
$$\frac{1}{(t+\alpha)(t+\beta)} = \frac{A}{t+\alpha} + \frac{B}{t+\beta}$$
After solving this I end up with some $A, B$ coefficients and I can solve the integral.
Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.
(bonus question: what if there were more variables, like 3 variables?)
I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.
 A: If your fraction is in form of $$ \frac {1}{(t-\alpha_1)(t-\alpha_2)(t-\alpha_3)...(t-\alpha_k)}$$ where the $\alpha s$ are different, there is an easy way to find the partial fraction decomposition. 
Let $t=\alpha_i$ and cover $(t-\alpha_i)$ while evaluating the above fraction to get your $A_i$
For example $$ \frac{1}{(t-1)(t-3)(t+4)} = \frac {A_1}{(t-1)} + \frac{ A_2}{(t-3)} +\frac { A_3}{(t+4)}$$
Where $$A_1 = \frac {1}{(1-3)(1+4)}=\frac{-1}{10}$$
$$A_2 = \frac {1}{(3-1)(3+4)} = \frac{1}{14}$$
$$A_3=\frac {1}{(-4-1)(-4-3)}=\frac{1}{35}$$ 
A: Unless I am misunderstanding the question, by "root rotation" (when $\beta \neq \alpha$):
$$\frac{1}{(t + \alpha)(t + \beta)} = \frac{A}{t + \alpha} + \frac{B}{t + \beta}$$
$$1 = A(t + \beta) + B(t + \alpha)$$
Evaluating $-\beta$ for $t$:
$$1 = B(\alpha - \beta)$$
$$B = \frac{1}{\alpha - \beta}$$
Similarly, for $A$, sub in $-\alpha$:
$$1 = A(\beta - \alpha)$$
$$A = \frac{1}{\beta - \alpha}$$
A: Here's your answer
for general $n$.
$\dfrac1{\prod_{k=1}^n (x-a_k)}
=\sum_{k=1}^n \dfrac{b_k}{x-a_k}
$.
Therefore
$1
=\sum_{k=1}^n \dfrac{b_k\prod_{j=1}^n (x-a_j)}{x-a_k}
=\sum_{k=1}^n b_k\prod_{j=1, j\ne k}^n (x-a_j)
$.
Setting
$x = a_i$
for each $i$,
all the terms 
except the one with $k=i$
have the factor $a_i-a_i$,
so
$1 = b_i\prod_{j=1, j\ne i}^n (a_i-a_j)
$
so that
$b_i
=\dfrac1{\prod_{j=1, j\ne i}^n (a_i-a_j)}
$.
For $n=2$,
$b_1
=\dfrac1{a_1-a_2}
$,
$b_2
=\dfrac1{a_2-a_1}
$.
For $n=3$,
$b_1
=\dfrac1{(a_1-a_2)(a_1-a_3)}
$,
$b_2
=\dfrac1{(a_2-a_1)(a_2-a_3)}
$,
$b_3
=\dfrac1{(a_3-a_1)(a_3-a_2)}
$.
