Integration by Parts, Taught the Way It Should Be

July 2026

Original post: @sonukg4india on X, 2026-07-21. If the embed above doesn’t load, everything it showed is written out below.

The rule itself

For two continuously differentiable functions $u(x)$ and $v(x)$:

$$\int u\,dv = uv - \int v\,du$$

That’s the whole rule. Everything else is bookkeeping for applying it well.

ILATE: how to actually choose which function is “$u$”

The single hardest part of integration by parts in practice isn’t the formula — it’s picking which factor to treat as $u$ and which as $dv$ when integrating a product $\int f(x)\,g(x)\,dx$. Pick wrong and the integral gets more complicated instead of less. ILATE is a real, working mnemonic for getting this right on the first try, in priority order for which function to choose as “$u$”:

1. Inverse (e.g. $\arcsin x$, $\arctan x$)
2. Logarithmic (e.g. $\log x$)
3. Algebraic (e.g. $x^2$, polynomials)
4. Trigonometric (e.g. $\sin x$, $\cos x$)
5. Exponential (e.g. $e^x$)

Whichever function appears earliest on this list becomes $u$; the rest becomes $dv$. The reasoning behind the order: functions higher on the list tend to get simpler when differentiated (an inverse or log function’s derivative is often algebraic), while functions lower on the list stay easy to integrate repeatedly (exponentials integrate right back into themselves) — so the order is built around which factor benefits more from being differentiated versus integrated.

Worked example: $\int \log(x)\,dx$

This is the example that makes ILATE click, because it looks like there’s only one function, not a product — the trick is recognizing $\log(x)$ as $\log(x)\cdot1$:

$$\int (\log x)\cdot 1\,dx = \log x\int 1\,dx - \int\left(\frac{d(\log x)}{dx}\int 1\,dx\right)dx$$

By ILATE, $\log x$ (Logarithmic) outranks the implicit $1$ (Algebraic), so $u=\log x$, $dv=1\,dx$:

$$= (\log x)\,x - \int\frac{1}{x}\cdot x\,dx = x\log x - \int 1\,dx = x\log x - x + C$$

Catalog status: Proven Systems

Integration by parts is a direct, rigorously provable consequence of the product rule for differentiation, run in reverse — real, foundational calculus, not a claim under test.

Why this is a good example of the section’s whole point

Nothing above is new mathematics. What makes it worth featuring is the teaching: a clean statement of the rule, a real memory aid for the one genuinely hard part (choosing $u$), and a worked example chosen specifically because it looks like it doesn’t fit the pattern until you see the trick. That’s what “simplified” should actually mean — not less rigorous, just taught by someone who remembers what was confusing about it the first time.