Integral of 1/(x^3+1) from 100 integrals

Integral of 1/(x^3+1) with partial fractions and completing the square!

blackpenredpen

[blackpenredpen]

If you have seen this one already,





thank you!

[EpicMathTime]

Physicist solution: assume that the + 1 is negligible so just integrate 1/x^3 instead.

[angelmendez-rivera]

This is generalizable to 1/(x^n + 1) for all natural n. Simply remember that the roots of x^n + 1 are all the nth roots of negative unity. Since every root that is not -1 or 1 will have a complex conjugate, these linear complex factors multiply in a trivial way to quadratic real irreducible factors of the form x^2 - 2·cos(2πm/n + π/n)·x + 1, where This means that the antiderivative will always be a linear combination of the logarithms of the linear real factors, which will always be x - 1 and x + 1, plus a linear combination of the logarithms of quadratic factors of the aforementioned form, plus a linear combination of the arctangents of factors of the form (x - cos(2πm/n + π/n))/sin(2πm/n + π/n) = csc(2πm/n + π/n)·x - cot(2πm/n + π/n). In this case, because we are using conjugates and the branch (-π, π) for complex logarithms, m runs from 0 to floor(n/2). Furthermore, we know that for all n, the coefficient of the vector ln|x - 1| in this case will always be 0. This gives a semi-trivial answer to the antiderivatives of 1/(x^n + 1).

All that remains is to find these coefficients as a function of n, which is quite a hardy task. The coefficient of ln|x + 1| is always the next easiest to find, since it is 0 for all even n. For odd n, you can multiply by x + 1, leaving the resulting coefficient as a constant, with the rest of the right side of the equation being multiplied by said factor, while the left side is (x + 1)/(x^n + 1) = 1/[x^(n - 1) - x^(n - 2) + ••• - x + 1] = 1/[(-x)^(n - 1) + (-x)^(n - 2) + ••• + (-x) + (-x)^0]. For x = -1, this equals 1/n, and the right side of the equation is simply equal to the desired constant, meaning that for odd n, the constant is equal 1/n, while for even n, it is 0. This can be expressed easily as [1 - (-1)^n]/(2n). This is what this constant will always equal for any n. The other constants are non-trivial, but because the natural numbers will never be an nth root of negative unity, you can always substitute 0, 1, and so on, into the resulting partial fraction decomposition pre-integration, and you will eventually get a system of linear equations. Solving this linear system gives you the coefficients of the linear combination. Done. Now, the only remaining challenge left to have a complete, closed-form solution is to transform this system of linear equations into a matrix linear equation which is methodically solvable for all n. This means we want to find the vector A, we know the vector (x^2 + stuff·x + 1)/(x^n + 1) with x = 0, 1, 2, 3, ..., M, which we call N, and we want to find the way to express the matrix M that transforms A to N as a function of n and its indeces. Then, once this is done, all one would need to do is to find a way to invert this square matrix, and the rest is relatively trivial.

With this plan of action, the first thing one would need to is find how many quadratic factors x^n + 1 has. The answer is floor(n/2), and this is easy to see, because the number of quadratic factors has to be doubled and then add 1 or 0, depending on the parity of n, to achieve degree n. This means there are n/2 factors if n is even, with no factor of x + 1, and (n - 1)/2 factors if n is odd, with the one factor of x + 1. floor(n/2) covers both cases. This also makes sense, since the quadratic factors have a linear term which is equal to the cosine of 2πm/n + φ, and m only runs to floor(n/2), as I stated earlier. With this known, we have to account for the fact that for each quadratic factor, there are two constants: the linear coefficient, and the additive constant. The additive constant always ends being a factor of the coefficient of an arctangent, while the linear constants are the coefficients of a logarithm of a quadratic factor. This means there are 2·floor(n/2) constants to solve for, which means the vector we want to solve for and parametrize has dimension 1 by 2·floor(n/2), and the square matrix we want to parametrize has dimension 2·floor(n/2) by 2·floor(n/2). This information should make a generalization method much easier.

[singfredabotane]

All of your videos are indeed helpful with clear explanation.Highly appreciate your time and commitment.

[ThAlEdison]

For irreducible quadratics in partial fractions, instead of multiplying by x, I multiply by the derivative of the partial fraction.
We can still use A=1/3, but the rest becomes

For x=0: 1=(1/3)-B+C, 2/3=-B+C
For x=1: 1/2=1/6+B+C, 1/3=B+C
Giving C=1/2 and B=-1/6
That let's me do the following:


More work during the partial fractions, less work when integrating.

[majkgmajkg]

My nightmare-intergral from studies. It took me 2 days to destroy it... Now it seems so easy... After few years it is nice to remind something like that. Thanks! :)

[KStarGamer_]

... 0:19 Did he just quote the "Number 15" Meme? lmao #yay

[jashwanth]

Thank you sir your content is very useful and .I promise that i only use you videos for my entire calculas preparation.❤❤

[janda]

This brings back memories from the integral of cube root - tan x

[rishabhanand]

Your first step to loving integrals starts by solving this magnificent beast

Integral(dx/((1+x^3)^3))

[abhishektyagi]

SIR THE RESOURCES AND LINKS TO LEARN MATHEMATICS THAT YOU SAID IN YOUR VIDEO WITH fematika ARE STILL NOT UPLOADED IN THE DESCRIPTION OF THE VIDEO, please do upload those links

[KM-omhm]

Without doing those

1=1-x²+x²

Now integration of 1-x²/(1+x³)
1-x/1-x²+x^4=

We know how to integrate

Now the remaining part
x/1-x²+x^4
x²=u

Done

[mdmustakalamalam]

Sir, your all videos are helpful for me so thanks sir.

[kirtithapar]

Amazing explanation sir, Thankew very much 🙏🏻❤️

[vartan_babayan_]

sorry, but I did not understand how and why did you use -1 and 1 and 0 to find A, B and C, seems like magic, can you explain or give some links where I can read about how to integrate using method of undefined factors much fastly, thank you :)

[srishtinegi]

Hello sir you teach so well
Made integral easiest for me
You belong to China ?

[akshayrajpoot]

Great sir 😘😘


Love from India ❤️🇮🇳🇮🇳🇮🇳

[plislegalineu]

Everyone's problems: "How do I get a job?" "How do I get a partner?"
Mathematicians' problems: "What's the integral of 1/(x³+1) dx?"
BPRP's problems: "What color to use?"

[idirbloodfallen]

Let's goo i I searched for this integral and it was a surprise that my favorite teacher made it

[carpediemyes]

How did you get that A = 1/3 again ?? That was not at all clear to me.