A person in another forum answered me with a method that I had never seen, I understand how it works except for the end where he gives me suggestions on how to solve it, it's been months and I still don't understand what I have to do to get the last ignconites.
Look at the answer of Shambhu Bhat https://www.quora.com/How-do-you-to-evaluate-the-integral-of-int-frac-1-3x-5-x-2-2-x-2-6-x-2-x-1-2-dx-Is-it-possible-to-do-any-tricks-before-or-in-the-partial-fractions-themselves-1
I tried to follow Shambhu's suggestions, but I only get this, I also tried with the other values but the result is the same. $ \begin{array}{l}\begin{array}{l}-------------------------------------------------\\ \left[\frac{6561}{3450799}=A\right]\left[\frac{?}{?}=B\right]\left[\frac{1}{5390}=C\right]\left[\frac{557}{3795950}=D\right]\left[-\frac{191}{3795950}=E\right]\left[\frac{?}{?}=F\right]\left[\frac{?}{?}=G\right]\left[\frac{36}{26269}=H\right]\left[\frac{185}{26269}=I\right]\\ -------------------------------------------------\end{array}\end{array} $
$ \frac{1}{\left(3x+5\right)\left(x-2\right)^2\left(x^2+6\right)\left(x^2+x+1\right)^2}=\frac{A}{\left(3x+5\right)}+\frac{B}{\left(x-2\right)}+\frac{C}{\left(x-2\right)^2}+\frac{Dx+E}{\left(x^2+6\right)}+\frac{Fx+G}{\left(x^2+x+1\right)}+\frac{Hx+I}{\left(x^2+x+1\right)^2} $
$\frac{1}{\left(3x+5\right)\left(x-2\right)^2\left(x^2+6\right)\left(x^2+x+1\right)}=\frac{A\left(x^2+x+1\right)}{\left(3x+5\right)}+\frac{B\left(x^2+x+1\right)}{\left(x-2\right)}+\frac{C\left(x^2+x+1\right)}{\left(x-2\right)^2}+\frac{\left(Dx+E\right)\left(x^2+x+1\right)}{\left(x^2+6\right)}+Fx+G+\frac{Hx+I}{\left(x^2+x+1\right)} $
$ \left[x=0\right] $
$ \frac{1}{\left(3\left(0\right)+5\right)\left(0-2\right)^2\left(0^2+6\right)\left(0^2+0+1\right)}=\frac{A\left(0^2+0+1\right)}{\left(3\left(0\right)+5\right)}+\frac{B\left(0^2+0+1\right)}{\left(0-2\right)}+\frac{C\left(0^2+0+1\right)}{\left(0-2\right)^2}+\frac{\left(D\left(0\right)+E\right)\left(0^2+0+1\right)}{\left(0^2+6\right)}+F\left(0\right)+G+\frac{H\left(0\right)+I}{\left(0^2+0+1\right)} $
$ \frac{1}{120}=\frac{A}{5}-\frac{B}{2}+\frac{C}{4}+\frac{E}{6}+G+I $
$ \frac{1}{120}=\frac{\frac{6561}{3450799}}{5}-\frac{B}{2}+\frac{\frac{1}{5390}}{4}+\frac{-\frac{191}{3795950}}{6}+G+\frac{185}{26269}\ ? $
there is nothing to compare there is not 1 single igconigta left, is anyone able to complete this person's method?
Let's look at a simpler example:
$$\frac{1}{(3x+5)(x-2)^2} = \frac{A}{3x+5} + \frac{B}{x-2} + \frac{C}{(x-2)^2},$$ or $$1 = (x-2)^2 A + (x-2)(3x+5)B + (3x+5)C.$$ Then if $x = 2$, all but the last term on the right hand side become zero, so $$1 = 11C,$$ hence $C = 1/11$. similarly, if $x = -5/3$, then all but the first term become zero, so $$1 = (-\tfrac{5}{3} - 2)^2 A = \frac{121}{9}A,$$ hence $A = \frac{9}{121}$. Unfortunately, there is no choice of $x$ such that the first and last terms simultaneously equal zero and let us solve for $B$. However, this is not difficult to overcome; we can just choose a convenient value, say $x = 1$, and substitute the known values of $A$ and $C$: $$1 = A - 8B + 8C = \frac{9}{121} - 8B + \frac{8}{11},$$ hence $B = -\frac{3}{121}$.
A similar idea applies to the more complicated rational function. The only issue is that $x^2 + 6$ and $x^2 + x + 1$ have no real roots; but because the equation must identically hold, there is nothing prohibiting us from selecting an appropriate complex-valued root; e.g., $x = i \sqrt{6}$ will work. For example:
$$\frac{1}{(3x+5)(x^2 + 6)} = \frac{A}{3x+5} + \frac{Dx + E}{x^2 + 6}$$ implies $$1 = (x^2 + 6)A + (3x+5)(Dx + E),$$ and the choice $x = -5/3$ this time gives $$1 = \frac{79}{9}A,$$ hence $A = \frac{9}{79}$. Now the choice $x = i \sqrt{6}$ yields $$1 = (3\sqrt{6} i + 5)(E + i \sqrt{6} D) = (5E - 18D) + (5 \sqrt{6} D + 3 \sqrt{6} E)i.$$ The imaginary part must be zero, so $5D + 3E = 0$, and the real part must be $1$, so $5E - 18D = 1$. This gives $D = -\frac{3}{79}$, $E = \frac{5}{79}$.
The key idea is to first solve for the easy coefficients--the ones that have a linear factor in the denominator. Then use these to solve for the higher-order terms.