From what I can see so far, I know that the error is early on in the work, at least as far as the inverse tangent term is concerned. The second term of the anti-derivative is only incorrect for the final stage, as mistakes had otherwise cancelled each other out. For the final answer, you multiplied the numerator and denominator by x^6, which would leave x^6/x^3 on top, for x^3 (you have x^2).

I find that using the substitution of x^3 initially is much easier to work with, but I have seen that the integrals are identical up to this point.

Additionally, I've noticed that using the substitution t=cot(theta) is able to bring back a success, as this would be the same as just using x^3 at the beginning.

I would like to know if there is any method to make an estimation on how big it is a number, in terms of orders of magnitude of n! when n is a big number.

(x^2) (2-x)^2 = 1 + 2(2-x)^2 I asked this problem on this page a few months ago, but didn’t get any satisfactory answer. I think there is a trick solution to it

From what I can see so far, I know that the error is early on in the work, at least as far as the inverse tangent term is concerned. The second term of the anti-derivative is only incorrect for the final stage, as mistakes had otherwise cancelled each other out. For the final answer, you multiplied the numerator and denominator by x^6, which would leave x^6/x^3 on top, for x^3 (you have x^2).

I find that using the substitution of x^3 initially is much easier to work with, but I have seen that the integrals are identical up to this point.

Additionally, I've noticed that using the substitution t=cot(theta) is able to bring back a success, as this would be the same as just using x^3 at the beginning.