A powerful integration trick.

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[Mystery_Biscuits]

Another method to solve these kinds of integrals is using the residue theorem. For integrands that are rational functions where the degree of the denominator is at least 2 more than the degree of the numerator, one can close the contour in the complex plane with a large semi-circle and then show (using Jordan's Lemma or similar) that it does not contribute in the limit. Hence, the integral is 2 pi i * sum of the residues of the (complexified) integrand in the upper half plane, and given that it's a rational function, that's not hard to compute.

In this case, note that the integrand is even, so we can say I = 1/2 * \int_{-\infty}^\infty (x^2+1)/(x^4+7x^2+1) dx.
Finding the zeros of the denominator:
x^2 =( -7\pm \sqrt{7^2 - 4})/2 = (-7\pm 3\sqrt{5})2. Taking positive square roots (as we only want singularities in the upper half plane) we have z_0 = i(3 - \sqrt{5})/2 and z_1 = i(3 + \sqrt{5})/2.
Clearly all singularities are simple poles so computing the residue is fairly short.
Res((z^2+1)/(z^4+7z^2+1), z_0) = (z_0^2+1)/(4z_0^3+14z_0) = -i/6 (Using the fact that if g/h has a simple pole at z_0 (and h'(z_0) \neq 0), then the residue is g(z_0)/h'(z_0).)
Res((z^2+1)/(z^4+7z^2+1), z_1) = (z_1^2+1)/(4z_1^3+14z_1) = - i/6.
Then I = 1/2 * (2 pi i) * (-i/6 + -i/6) = pi/3.

[mathunt]

This only works for very specific integrals.

[oo_rf_oo]

This type of integral (by that I mean the integrand is (dx^2+e)/(ax^4+bx^2+c) and the denominater has real no roots) has a nice closed form, it's In this question specifically, a=c=d=e=1, b=7, plug the numbers in we have π/3.

[mathisnotforthefaintofheart]

This method is more of a "hack", which is a nice one by the way, I admit. Notice, that if the 1 in the numerator becomes an x, this method no longer works.

[ArminVollmer]

Hey Michael, good work as always! If I could wish something, then possibly more functional equations, or rather still examples to Polya's enumeration method (Isomer counting principle using group theory). Thanks!

[bobzarnke]

I wondered about the more general case of ∫[0, ∞] (x²+1)/(x⁴+ax²+b) dx, using the conventional approach of factoring the denominator, separating into partial
fractions and integrating. This gives π(1+√b)/(2√b√(a+2√b)). When b=1, as above, this is π/√(a+2).

[fafablablabla]

I just had calculus exam yesterday LOL

[richardheiville]

For generalization see G. Boole theorem or the master formula of Ramanujan.

[sircombos]

I'm confused by 2:40, how did he get the denominator to be 9 + (x +1/x)? He is dividing the denominator by x^2 so surely it would be 9 + x^2 -2 +1x^-2, I can't see where he managed to get (x+1/x). Could anyone explain this better?

[stanschmidt]

would you like doing a Video on what will result in?

[ecoidea]

took me like 3 hours but I made, first with substitution u=tan(x), then (sin²(x)+cos²(x))²=1, then back again with tan(x)=u

[jackball]

Would love to see you and Professor Leonard go head to head in an arm-wrestling match.

[DD-ybht]

1 + 1/x^2 is the derivative of x -1/x.

[AtariDays]

Just curious, at 5:13, can't you go a step further and say its 4x integral from 0 to 1? I mean, area from 0 to 1 is same as 1 to infinity, no?

[DanielCastro-uylu]

1/(x^5+1) is a classic, even 1/(x^4+1) or 1/(x^6+1)
But what about 1/(x^7+1) or the general case 1/(x^n+1) 🤯
But not definite integral, just the antiderivate

[Anti_Electron]

can someone explain to me why after intergrating from minus infinity to positive infinity he went to 2 intergrate from 0 to positive infinity

[suupk]

I would be really cool if you could generalize these and show them at the end. I think that would be something interesting.

[enthusiast]

I think that isn't a trick but the standard and only method to solve such problems

[loich.]

I don't understand why you can devide bue x² because x=0 is the lower bound...

[johns.]

So how do you solve this if you have x^2+2 in the numerator instead of x^2+1 ?