Otdfbt

Often I am dealing with an integral of let's say:

$$intfracdt(t-2)(t+3)$$

or

$$int fracdtt(t-4)$$

or to make this a more general case in which I am interested the most:

$$int fracdt(t+alpha)(t+beta) quad quad alpha, beta in mathbbR$$

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):

$$frac1(t+alpha)(t+beta) = fracAt+alpha + fracBt+beta$$

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.

Here's your answer
for general $n$.

$dfrac1prod_k=1^n (x-a_k)
=sum_k=1^n dfracb_kx-a_k
$.

Therefore
$1
=sum_k=1^n dfracb_kprod_j=1^n (x-a_j)x-a_k
=sum_k=1^n b_kprod_j=1, jne 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_iprod_j=1, jne i^n (a_i-a_j)
$
so that
$b_i
=dfrac1prod_j=1, jne i^n (a_i-a_j)
$.

For $n=2$,
$b_1
=dfrac1a_1-a_2
$,
$b_2
=dfrac1a_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)
$.

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 $$ frac1(t-1)(t-3)(t+4) = frac A_1(t-1) + frac A_2(t-3) +frac A_3(t+4)$$
Where $$A_1 = frac 1(1-3)(1+4)=frac-110$$
$$A_2 = frac 1(3-1)(3+4) = frac114$$

$$A_3=frac 1(-4-1)(-4-3)=frac135$$

Unless I am misunderstanding the question, by "root rotation" (when $beta neq alpha$):

$$frac1(t + alpha)(t + beta) = fracAt + alpha + fracBt + beta$$

$$1 = A(t + beta) + B(t + alpha)$$

Evaluating $-beta$ for $t$:

$$1 = B(alpha - beta)$$

$$B = frac1alpha - beta$$

Similarly, for $A$, sub in $-alpha$:

$$1 = A(beta - alpha)$$

$$A = frac1beta - alpha$$

Let $n$,$m$ be integers such that $nge 2$ and $n-1 ge m ge 0$ .Then if the roots of the polynomial in question are all distinct the following formula holds:
beginequation
fracx^mprodlimits_j=1^n (x-b_j) = sumlimits_i=1^n frac1x-b_i cdot fracb_i^mprodlimits_j=1,jneq i^n (b_i-b_j)
endequation
Now, if the roots are not distinct then it gets more complicated but even at the outset it is clear that we can generate a similar formula by differentiating both sides with respect to the particular root a certain number of times. For example if $n=2$ and $m_1 ge 1$, $m_2 ge 1$ and $mge 0$ the following formula holds true:
begineqnarray
fracx^m(x-b_1)^m_1 (x-b_2)^m_2=sumlimits_j=0^m left( sumlimits_l_1=1^m_1 binomm_1+m_2-1-l_1m_2-1(-1)^m_2 b_1^j frac1_l_1=1 binomm-1j-1 + 1_l_1>1 binommj(x-b_1)^l_1-m+j (-b_1+b_2)^m_1+m_2-l_1+
sumlimits_l_1=1^m_2 binomm_1+m_2-1-l_1m_1-1(-1)^m_1 b_2^j
frac1_l_1=1 binomm-1j-1 + 1_l_1>1 binommj(x-b_2)^l_1-m+j (-b_2+b_1)^m_1+m_2-l_1
right)
endeqnarray

In[5732]:= ll = ; x =.; b1 =.; b2 =.; m = RandomInteger[0, 10];
For[count = 1, count <= 100, count++,
 m1, m2 = RandomInteger[1, 5, 2]; x =.; b1 =.; b2 =.;
 xx1 =
 Sum[ 
 Binomial[m1 + m2 - 1 - l1, m2 - 1] ((-1)^
 m2 b1^j If[l1 == 1, Binomial[m - 1, j - 1], 
 Binomial[m, j]])/((x - b1)^(l1 - m + j) (-b1 + b2)^(
 m1 + m2 - l1)), l1, 1, m1, j, 0, m] + 
 Sum[ Binomial[m1 + m2 - 1 - l1, m1 - 1] ((-1)^
 m1 b2^j If[l1 == 1, Binomial[m - 1, j - 1], 
 Binomial[m, j]])/((x - b2)^(l1 - m + j) (-b2 + b1)^(
 m1 + m2 - l1)), l1, 1, m2, j, 0, m];
 xx2 = x^m/((x - b1)^m1 (x - b2)^m2);
 ll = Join[ll, Simplify[xx1 - xx2]];
 ];
ll

Out[5734]= 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

It would be interesting to derive similar formulae in the case $n > 2$.