When U-Substitution Fails

U-substitution handles composites: if you see f(g(x))·g'(x), let u = g(x) and you're done. But ∫x·eˣ dx has no composition hiding in it. x is not the derivative of eˣ, and eˣ is not the derivative of x. There is no u that makes both disappear.

The integrand is a product of two unrelated functions, not a composite. That is when integration by parts enters.

The Formula and Where It Comes From

The Product Rule says d(uv)/dx = u·v' + v·u'. Integrate both sides:

uv = ∫u dv + ∫v du

Rearrange: ∫u dv = uv − ∫v du. That is integration by parts. It does not solve the integral — it transforms it into a different integral (∫v du) that you hope is easier. The whole skill is choosing u and dv so that ∫v du is simpler than what you started with.

Choosing u — The LIATE Guide

Choose u from whichever function type comes first in this list:

L Logarithms — ln x, log x I Inverse trig — arctan, arcsin A Algebraic — xⁿ, polynomials T Trigonometric — sin, cos E Exponential — eˣ

The reason this ordering works: functions at the top of the list (logs, inverse trig) simplify drastically when differentiated — ln x becomes 1/x, arctan x becomes 1/(1+x²). Functions at the bottom (trig, exponential) just cycle when differentiated or integrated, so you want to put those in dv.

LIATE is a guideline, not a law. It breaks down occasionally, particularly when both functions are the same type. When it breaks down, try both assignments and see which gives a simpler ∫v du.

Four Worked Examples

📋 ∫x·eˣ dx — the standard case
Chooseu = x (algebraic, A in LIATE), dv = eˣ dx (exponential, E)
Computedu = dx, v = eˣ
Applyx·eˣ − ∫eˣ dx = x·eˣ − eˣ + C
Answereˣ(x − 1) + C
Verifyd/dx[eˣ(x−1)] = eˣ(x−1) + eˣ = xeˣ ✓
📋 ∫ln(x) dx — just a log with no obvious partner
TrickWrite as ∫ln(x)·1 dx. Set u = ln x (L), dv = 1·dx.
Computedu = (1/x)dx, v = x
Applyx·ln(x) − ∫x·(1/x)dx = x·ln(x) − ∫1 dx
Answerx·ln(x) − x + C
📋 ∫x²·sin(x) dx — needs two rounds
Round 1u = x², dv = sin x dx. du = 2x dx, v = −cos x. Result: −x²cos x + ∫2x cos x dx
Round 2 on ∫2x cos x dxu = 2x, dv = cos x dx. du = 2dx, v = sin x. Result: 2x·sin x − ∫2 sin x dx = 2x·sin x + 2cos x
Combine−x²cos x + 2x·sin x + 2cos x + C
NoteFor ∫xⁿ times trig/exp, you need n applications. The tabular method saves time here.
📋 ∫eˣ·sin(x) dx — the cyclic case
Round 1u = sin x, dv = eˣ dx. Result: eˣ·sin x − ∫eˣ·cos x dx
Round 2u = cos x, dv = eˣ dx. Result: eˣ·sin x − [eˣ·cos x + ∫eˣ·sin x dx]
Spot itThe original integral I has reappeared on the right: I = eˣ·sin x − eˣ·cos x − I
Solve2I = eˣ(sin x − cos x) → I = eˣ(sin x − cos x)/2 + C

The Tabular Method for Repeated IBP

When the polynomial factor is high degree — like ∫x⁴·eˣ dx needing four rounds — the tabular method reduces the whole computation to a table:

📋 ∫x³·cos(x) dx via tabular method
Derivatives column (u)x³ → 3x² → 6x → 6 → 0
Integrals column (dv)cos x → sin x → −cos x → −sin x
Signs (+−+−)Multiply diagonally with alternating signs
Answerx³·sin x + 3x²·cos x − 6x·sin x − 6cos x + C

Where Students Go Wrong

Forgetting the minus sign. ∫u dv = uv minus ∫v du. Students write "plus" half the time. Check your sign before doing anything else.

Choosing u and dv backwards. If you pick u = eˣ and dv = x dx for ∫x·eˣ dx, you get ∫v du = ∫(x²/2)·eˣ dx — harder than what you started with. The error is obvious in hindsight: always check that ∫v du looks simpler before committing.

Applying IBP when u-sub would work. ∫x·e^(x²) dx is a u-substitution (u = x²), not IBP. The presence of x alongside e^(x²) is the clue — x is the derivative of x², so the chain rule reversal works.

Frequently Asked Questions

LIATE says to put ln first, but why not eˣ first?▾
If you set u = eˣ and dv = ln(x)dx, you need to integrate ln(x) to find v — which requires IBP itself. You get a circular problem. Setting u = ln(x) and differentiating gives the clean du = (1/x)dx, avoiding the problem entirely. The ordering in LIATE is specifically designed to put hard-to-integrate functions in u (where you differentiate) and easy-to-integrate functions in dv.
When does the "cyclic" pattern happen?▾
When both factors loop under differentiation/integration — typically exponential times trig. eˣ is its own derivative, and differentiating sin gives cos which gives −sin which cycles back. After two rounds of IBP you have the original integral on both sides of the equation — solve algebraically for I rather than applying IBP a third time.
How do I know when to use IBP versus u-substitution?▾
U-substitution: you see a composite function and its derivative lurking nearby. IBP: you see a product of two unrelated functions — a polynomial times something, or a logarithm with no obvious substitution. The tell for IBP is that differentiating one factor simplifies it while the other integrates without getting worse.