Integral by Parts Formula- Techniques and Examples
What Integration by Parts Actually Is
Integration by parts is a technique that transforms integrals of products into something easier to handle. It comes from the product rule for differentiation, reversed.
Most students memorize the formula without understanding when to use it. That's where they go wrong. You need to recognize specific function types before you even attempt this method.
The Formula
Here it is:
∫ u dv = uv − ∫ v du
That's it. The entire method rests on this one equation. You pick one part of your integrand to be u and the other part to be dv. Then you differentiate u to get du and integrate dv to get v.
The trick is choosing wisely. Pick wrong and you'll make the integral harder instead of easier.
The LIATE Rule — Your Decision Framework
Most textbooks teach LIATE. It works in most cases. Here's the hierarchy:
| Letter | Type | Example |
|---|---|---|
| L | Logarithmic | ln(x) |
| I | Inverse trig | arctan(x) |
| A | Algebraic | x², x³ |
| T | Trigonometric | sin(x), cos(x) |
| E | Exponential | eˣ, 2ˣ |
Pick u from earlier in the list. Choose dv from later.
⚠️ Warning: LIATE isn't foolproof. For ∫ x·ln(x) dx, it says let u = ln(x) and dv = x dx. That's correct. But for ∫ x·eˣ dx, some argue u = x and dv = eˣ dx works better. Use your judgment.
How to Actually Do It — Step by Step
Here's the process:
- Identify two factors in your integrand
- Apply LIATE to decide which is u and which is dv
- Compute du by differentiating u
- Compute v by integrating dv
- Plug everything into ∫ u dv = uv − ∫ v du
- Solve the remaining integral
Sometimes you need to apply integration by parts twice. Sometimes you get the original integral back — that's when you solve algebraically.
Examples That Actually Teach You Something
Example 1: ∫ x·eˣ dx
This is the textbook example. LIATE says: u = x (algebraic, earlier in the list) and dv = eˣ dx (exponential, later).
Step 1: u = x, so du = dx
Step 2: dv = eˣ dx, so v = eˣ
Step 3: Apply the formula
∫ x·eˣ dx = x·eˣ − ∫ eˣ dx
Step 4: Solve the remaining integral
∫ eˣ dx = eˣ
Final answer: x·eˣ − eˣ + C
Factor if you want: eˣ(x − 1) + C
Example 2: ∫ x²·ln(x) dx
LIATE puts ln(x) first, so u = ln(x) and dv = x² dx.
u = ln(x), so du = 1/x dx
dv = x² dx, so v = x³/3
Apply the formula:
∫ x²·ln(x) dx = (x³/3)·ln(x) − ∫ (x³/3)·(1/x) dx
Simplify the remaining integral:
∫ (x³/3)·(1/x) dx = ∫ x²/3 dx = x³/9
Final answer: (x³/3)·ln(x) − x³/9 + C
Example 3: ∫ eˣ·sin(x) dx
This one requires applying integration by parts twice, then solving algebraically.
First pass: Let u = sin(x), dv = eˣ dx
u = sin(x), du = cos(x) dx
v = eˣ
Result: ∫ eˣ·sin(x) dx = eˣ·sin(x) − ∫ eˣ·cos(x) dx
Second pass: Now you need ∫ eˣ·cos(x) dx. Let u = cos(x), dv = eˣ dx
u = cos(x), du = −sin(x) dx
v = eˣ
Result: ∫ eˣ·cos(x) dx = eˣ·cos(x) − ∫ −eˣ·sin(x) dx
∫ eˣ·cos(x) dx = eˣ·cos(x) + ∫ eˣ·sin(x) dx
Substitute back into the first result:
∫ eˣ·sin(x) dx = eˣ·sin(x) − [eˣ·cos(x) + ∫ eˣ·sin(x) dx]
∫ eˣ·sin(x) dx = eˣ·sin(x) − eˣ·cos(x) − ∫ eˣ·sin(x) dx
Move the remaining integral to the left side:
2∫ eˣ·sin(x) dx = eˣ·sin(x) − eˣ·cos(x)
Final answer: (eˣ/2)[sin(x) − cos(x)] + C
Example 4: ∫ ln(x) dx
This is a special case. You have only one function, but you can write ln(x) = 1·ln(x).
Let u = ln(x), dv = 1 dx
u = ln(x), du = 1/x dx
v = x
∫ ln(x) dx = x·ln(x) − ∫ x·(1/x) dx
∫ ln(x) dx = x·ln(x) − ∫ 1 dx
Final answer: x·ln(x) − x + C
Common Mistakes That Will Cost You Points
- Forgetting the constant of integration. Always add +C at the end.
- Picking u and dv backwards. This makes the integral harder. If that happens, stop and reverse your choices.
- Dropping the negative sign. du comes from differentiation — watch your signs carefully.
- Not simplifying between steps. Simplify the remaining integral before trying to solve it.
- Giving up after one application. Some integrals require two or more passes.
When Integration by Parts Is the Wrong Tool
Not every integral needs this method. Know what to use instead:
- u-substitution: For composite functions like ∫ 2x·eˣ² dx
- Trig substitution: For √(a² − x²), √(a² + x²), or √(x² − a²)
- Partial fractions: For rational functions where degree of numerator < degree of denominator
- Tabular integration: For ∫ polynomial × trig or ∫ polynomial × exp — faster than repeated parts
The Tabular Method — Faster for Specific Cases
When you have polynomial × (trig or exp), tabular integration beats repeated parts.
Make two columns. Left column: derivatives of polynomial until you hit 0. Right column: antiderivatives of the other function.
Multiply diagonally, alternating signs. Add the results.
Example for ∫ x²·eˣ dx:
| Differentiate (polynomial) | Integrate (eˣ) |
|---|---|
| x² | eˣ |
| 2x | eˣ |
| 2 | eˣ |
| 0 | eˣ |
Result: x²·eˣ − 2x·eˣ + 2·eˣ + C
Factor if you want: eˣ(x² − 2x + 2) + C
Bottom Line
Integration by parts isn't complicated. Pick u using LIATE, compute du and v, plug into uv − ∫v du, and solve what's left. Practice the basic cases until the pattern clicks. The tricky ones — like eˣ·sin(x) — just require patience and algebraic persistence.