Ideas from LLMs (sad, but still interesting).

Today’s suspect

The integral that we will be looking at is Daily Integral’s 9th Oct 2026 Beginner Integral

∫𝑒1𝑥2ln𝑥d𝑥

When I first saw this, a thought went through my mind: “integration by parts in beginner?” Things seem to be getting stale in the beginner section, and yet, IBP was very easy to spot in this case. So, instead I wondered what other creative ways are there to solve this seemingly benign integral.

Feynman?

Those who are familiar with “Feynman integration” or “Differentiation under the integral sign” may notice we have the ln𝑥 factor poking out, begging for Feynman’s trick to be used. So first, we parameterize the 2:

𝐹(𝛼)=∫𝑒1𝑥𝛼d𝑥=𝑒𝛼+1−1𝛼+1

Then differentiate our new function with respect to α:

𝐹′(𝛼)=dd𝛼∫𝑒1𝑥𝛼d𝑥=∫𝑒1𝜕𝜕𝛼𝑥𝛼d𝑥=∫𝑒1𝑥𝛼ln𝑥d𝑥

We also have another way to write 𝐹′(𝛼):

𝐹′(𝛼)=dd𝛼𝑒𝛼+1−1𝛼+1=𝛼𝑒𝛼+1+1(𝛼+1)2

Putting it all together:

∫𝑒1𝑥2ln𝑥d𝑥=𝐹′(2)=2𝑒3+19.

Double trouble

If you recall, natural logarithm can be defined by an integral:

ln𝑥=∫𝑥11𝑡d𝑡

So we will substitute that into our integral!

∫𝑒1𝑥2ln𝑥d𝑥=∫𝑒1𝑥2∫𝑥11𝑡d𝑡d𝑥

Now, let’s swap the order of the integral. We have 1≤𝑡≤𝑥≤𝑒, so the integral becomes

∫𝑒1∫𝑒𝑡𝑥2𝑡d𝑥d𝑡=∫𝑒1𝑒3−𝑡33𝑡d𝑡=𝑒33∫𝑒11𝑡d𝑡−13∫𝑒1𝑡2d𝑡=𝑒33−13(𝑒33−13)=2𝑒3+19.

Going exterior

The previous method relied on calculating the interior weighted area of a triangle. But with Green’s theorem —a special case of Generalized Stokes theorem, I just have to say it :)—we can compute path integrals instead

∬𝐷(𝜕𝑄𝜕𝑥−𝜕𝑃𝜕𝑦)d𝑥d𝑦=∮𝜕𝐷𝑃d𝑥+𝑄d𝑦

Choosing 𝑃=0,𝑄=𝑥33𝑦⟶𝑄𝑥=𝑥2𝑦 (and renaming 𝑡 to 𝑦) gives us

∫𝑒1∫𝑒𝑦𝑥2𝑦d𝑥d𝑦=∮𝑇𝑥33𝑦d𝑦

Where 𝑇 is the triangle with vertices (1,1),(𝑒,1),(𝑒,𝑒). Analyzing the edges, and their directions we get

∮𝑇𝑥33𝑦d𝑦=∫11𝑥33𝑦d𝑦⏟(1,1)→(𝑒,1)+∫𝑒1𝑒33𝑦d𝑦⏟(𝑒,1)→(𝑒,𝑒)+∫1𝑒𝑠33𝑠d𝑠⏟(𝑒,𝑒)→(1,1)𝑥=𝑦=𝑠=0+𝑒33+1−𝑒39=2𝑒3+19.

Differential algebra

Though I admit I don’t know much about this field, the suggested proof seems easy enough to understand, at least on the surface level. So, I will be sharing it here.

Consider the differential field 𝐾=ℂ(𝑥,ln𝑥) equipped with derivation 𝐷(𝑥)=1,𝐷(ln𝑥)=1𝑥. The original integrand 𝑥2ln𝑥 lies in the subspace 𝐶[𝑥]ln𝑥+𝐶[𝑥]. Let 𝐹(𝑥) be a “guess” or (ansatz) for 𝐷(𝐹)=𝑥2ln𝑥; it should have the form

𝐹(𝑥)=𝑝(𝑥)ln𝑥+𝑞(𝑥)⟹𝐷(𝐹)=𝑝′(𝑥)ln𝑥+𝑝(𝑥)𝑥+𝑞′(𝑥)

This gives us

𝑝′(𝑥)=𝑥2,𝑝(𝑥)𝑥+𝑞′(𝑥)=0

Note that a 𝑝(𝑥)𝑥+𝑞′(𝑥) has to be a polynomial, meaning the additive constant of integration of 𝑝(𝑥) has to be zero. Therefore, 𝑝(𝑥)=𝑥33,𝑞(𝑥)=−𝑥39+𝐶, and

𝐹(𝑥)=𝑥33ln𝑥−𝑥39+𝐶

Since 𝐹′(𝑥)=𝑥2ln𝑥 , by the fundamental theorem of calculus,

∫𝑒1𝑥2ln𝑥d=𝐹(𝑒)−𝐹(1)=(𝑒33ln𝑒−𝑒39)−(133ln1−139)=2𝑒3+19.

Conclusion

Integration is, itself, an art, the art that I still find difficult to truly wield. There are still plenty of unconventional ways to integrate ∫𝑒1𝑥2ln𝑥d𝑥. Sadly however, I do not yet have the knowledge required to fully grasp them. Maybe another time!