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.
Best Answer
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 $$A_1 = \frac {1}{(1-3)(1+4)}=\frac{-1}{10}$$ $$A_2 = \frac {1}{(3-1)(3+4)} = \frac{1}{14}$$
$$A_3=\frac {1}{(-4-1)(-4-3)}=\frac{1}{35}$$