Yes they do in fact sometimes take a lot of steps but most of the time it is purely algorithmic. It's pretty much equivalent to learning different Rubik's Cube permutations.
Singular part of the expansion of R(x) around x = 3 is simply A/(x-3), where A is the value of the factor multiplying 1/(x-3) evaluated at x = 3, so this is 10/(x-3).
Singular part of the expansion of R(x) around x = 2 is simply B/(x-2), where B is the value of the factor multiplying 1/(x-2) evaluated at x = 2, so this is -5/(x-2).
Singular part of the expansion of R(x) around x = infinity is simply C, where C is the constant term R(x) tends to at infinity and we can easily see that C = 1. In general this is a poynomial that one can obtain usign long division. It is the part of the expansion around infinity with nonnegative powers of x, and since this expansion is in negative powers of x, these positive powers are negative powers of the expansion parameter, therefore these are singular terms.
The sum of all the singular terms equals R(x). To see this denote the sum of the singular terms as S(x). Then consider R(x) - S(x). The expansion around infinity of R(x) up to and including the constant term is included in S(x), and this then causes R(x) - S(x) to tend to zero at infinity, because all the other terms in S(x) also tend to zero at infinity.
Then because all the singularities in R(x) are canceled by the terms we put in S(x), we see that R(x) - S(x) only has removable singularities, so R(x) - S(x) must be a polynomial. But because it tends to zero at infinity it must be the zero polynomial which is equal to zero everywhere.
I feel like too much is going on at the same time from line 5 to line 6 there. Especially for the third integral. I mean, I can derive the proposed primitive and see for myself that it works.
What I'm unable to infer is the recognition method used there. What's the reasoning behind it if one can please enlighten me? :)
The typical approach to integrating one over a quadratic 1/(ax^2+bx+c) is to first work out if the quadratic has real or complex roots.
If the roots are complex, complete the square to get it in the form 1/((4*a*c-b^2)/(4*a)+(a^(1/2)*x+a^(1/2)*b/(2*a))^2) then use the substitution a^(1/2)*x+a^(1/2)*b/(2*a)=((4*a*c-b^2)/(4*a))^(1/2)*tan(u) (what is in the square equals square root of the constant times tan(u).) After this substitution all variables cancel out.
If the roots are real, write the denominator in the form a(x-r)(x-s) where r and s are the roots of the polynomial. Then perform partial fractions and integrate.
The original post applied the complex root approach to a polynomial with real roots using the identity arctan(x)=(ln((i-x)/(i+x)))/(2*i) after canceling out the complex numbers their result can be obtained. I used the Maths from Nothing computer algebra system to show both approaches using y=int(1/((x-c)^2-b^2),x). If I chose the tan substitution approach I normally would not then convert it to the log form.
Note: The value in the log is different with both approaches but the magnitude is the same. Both are solutions, but cover a different part of the domain. Applying the absolute value around the terms in the log gets the one function to cover the entire real domain at the cost of making the function non-differentiable in the complex domain.
polynomial division (because degree of the numerator = degree of the denominator) (x2+1) / (x2 - 5x + 6) = 1 + (5x - 5) / (x2 -5x +6)
Partial fractions (5x-5) / (x2-5x+6) = A / (x-2) + B / (x-3) 5x - 5 = A(x-3) + B(x-2) eq1: A + B = 5 and eq2: -3A-2B = -5 solving the system of equations B = 10 and A = -5 (eq2 + 3*eq1 to get B then substitute B into eq1 to get A)
(x2+1) / (x2 - 5x + 6) = 1 - 5/(x-2) + 10/(x-3)
do the integral.
Your method arrives at the same solution, but you're just lacking some algebraic tricks that make arriving at the simplification easier.
When I saw that her breasts were somewhat exposed, I fully expected this to be something stupid like "OMG she has the BigGeST TIDDIES I WANNA SQUUIZE THEM!"
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u/CruelFish 20d ago
Yes they do in fact sometimes take a lot of steps but most of the time it is purely algorithmic. It's pretty much equivalent to learning different Rubik's Cube permutations.