Integration by Parts- Mastering the Method Step-by-Step
What Integration by Parts Actually Is
Integration by parts is a technique for breaking down complicated integrals into simpler pieces. It's the reverse of the product rule for derivatives. If you've got a product inside an integral and basic substitution won't work, this is your next move.
The formula comes from the product rule:
d(uv) = u dv + v du
Integrate both sides and rearrange, and you get:
∫u dv = uv - ∫v du
That's it. That's the whole method. You pick one part to call u and one part to call dv, then apply the formula.
The Hard Part: Choosing u and dv
Most students mess this up here, not in the algebra. Pick wrong and you'll end up with something harder than what you started with. Pick right and the integral practically solves itself.
Use the LIATE Rule
This mnemonic ranks function types by priority. When in doubt, pick from the top of the list:
- L - Logarithmic: ln(x)
- I - Inverse trig: arctan(x), arcsin(x)
- A - Algebraic: x², x³, constants
- T - Trigonometric: sin(x), cos(x)
- E - Exponential: eˣ, 2ˣ
Higher on the list gets u. The rest becomes dv.
Example: ∫x·eˣ dx → u = x (algebraic), dv = eˣ dx (exponential)
Example: ∫ln(x) dx → u = ln(x), dv = dx
Step-by-Step: Your First Integration by Parts
Let's work through ∫x·eˣ dx together.
Step 1: Identify u and dv
Using LIATE: x is algebraic, eˣ is exponential. Algebraic wins, so u = x and dv = eˣ dx.
Step 2: Find du and v
Take derivatives and antiderivatives:
- u = x → du = dx
- dv = eˣ dx → v = eˣ
Step 3: Plug into the Formula
∫u dv = uv - ∫v du
∫x·eˣ dx = x·eˣ - ∫eˣ·dx
∫x·eˣ dx = x·eˣ - eˣ + C
∫x·eˣ dx = eˣ(x - 1) + C
Done. That simple.
When You Need Two Rounds
Sometimes one application doesn't cut it. The remaining integral still has a product. You apply integration by parts again.
Example: ∫x²·eˣ dx
First round: u = x², dv = eˣ dx
= x²eˣ - ∫2x·eˣ dx
Second round on ∫x·eˣ dx: u = x, dv = eˣ dx
= x²eˣ - 2(xeˣ - eˣ) + C
= x²eˣ - 2xeˣ + 2eˣ + C
= eˣ(x² - 2x + 2) + C
Each application reduces the polynomial power by one. Keep going until it's gone.
Tabular Method: Faster for Repeated Rounds
When you have a polynomial times an exponential or trig function, the tabular method saves time. Derivatives on one side, antiderivatives on the other, alternate signs.
For ∫x²eˣ dx, build this table:
| Sign | D (derivatives of x²) | I (antiderivatives of eˣ) |
|---|---|---|
| + | x² | eˣ |
| - | 2x | eˣ |
| + | 2 | eˣ |
| - | 0 | eˣ |
Multiply down the diagonals with the signs:
= +x²eˣ - 2xeˣ + 2eˣ + C
Same answer, less thinking required.
Common Mistakes to Dodge
- Wrong u choice: If your new integral looks worse, restart with different choices. LIATE isn't perfect but it works most of the time.
- Dropping the constant: When you find v from dv, don't forget the +C thinking it'll show up later. It won't.
- Forgetting the minus sign: The formula has a minus before the second integral. People lose it under pressure.
- Not distributing: uv - ∫v du is not the same as uv - v·u. The second integral has its own variable.
Integration Techniques Comparison
| Technique | Use When | Key Requirement |
|---|---|---|
| Substitution | Composite function present | Chain rule structure |
| Integration by Parts | Product of different function types | LIATE applies |
| Trig Substitution | Square roots of a²-x², x²+a², x²-a² | Right triangle geometry |
| Partial Fractions | Rational function, degree num < deg denom | Factorable denominator |
Practice Problems to Try
Work through these before checking answers:
- ∫ln(x) dx
- ∫x·sin(x) dx
- ∫x²·ln(x) dx
- ∫eˣ·cos(x) dx
- ∫arcsin(x) dx
Answers: (1) xln(x) - x + C, (2) sin(x) - x·cos(x) + C, (3) x³/3·ln(x) - x³/9 + C, (4) eˣ(sin(x) + cos(x))/2 + C, (5) x·arcsin(x) + √(1-x²) + C
The Bottom Line
Integration by parts is mechanical once you internalize the formula and stop second-guessing your u choice. LIATE tells you what to pick. The formula tells you what to do. Practice until choosing u becomes automatic.