Integration by Parts- Formula and Worked Examples
What Is Integration by Parts?
Integration by parts is a technique that transforms an integral into a potentially simpler one. It comes from the product rule for differentiation, run backward.
You use it when you have a product of two functions and regular substitution won't work. That's it. If your integral looks like ∫f(x)g(x)dx and you can't find an antiderivative directly, this is your next move.
The Formula
The integration by parts formula is:
∫u dv = uv - ∫v du
Where:
- u is a function you differentiate
- dv is a function you integrate
The entire goal is to pick u and dv strategically so the new integral ∫v du is easier than what you started with.
When to Use This Method
You'll reach for integration by parts in these situations:
- Product of a polynomial and an exponential function
- Product of a polynomial and a logarithm
- Product of a polynomial and a trigonometric function
- Logarithmic functions on their own
- Inverse trigonometric functions
- Exponential functions
If you try substitution and it fails, try this next.
How to Choose u and dv (LIATE Rule)
Choosing wrong makes the problem harder. The LIATE rule gives you an order to try:
- L - Logarithmic functions (ln x, log x)
- I - Inverse trigonometric functions (arcsin x, arctan x)
- A - Algebraic functions (polynomials like x², x³)
- T - Trigonometric functions (sin x, cos x)
- E - Exponential functions (eˣ, 2ˣ)
Pick u from earlier in the list, and dv from later. This isn't foolproof, but it works for most textbook problems.
Worked Examples
Example 1: ∫x·eˣ dx
Step 1: Apply LIATE. x is algebraic, eˣ is exponential. So u = x, dv = eˣ dx.
Step 2: Find du and v.
- u = x → du = dx
- dv = eˣ dx → v = eˣ
Step 3: Plug into the formula.
∫x·eˣ dx = x·eˣ - ∫eˣ dx
Step 4: Solve the remaining integral.
∫eˣ dx = eˣ
Final answer: x·eˣ - eˣ + C
You can factor out eˣ: eˣ(x - 1) + C
Example 2: ∫x² ln x dx
Step 1: LIATE says ln x comes before x². So u = ln x, dv = x² dx.
Step 2: Find du and v.
- u = ln x → du = 1/x dx
- dv = x² dx → v = x³/3
Step 3: Apply the formula.
∫x² ln x dx = (ln x)(x³/3) - ∫(x³/3)(1/x) dx
Step 4: Simplify the second 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 a trick. You'll apply integration by parts twice.
First pass:
- u = sin x, dv = eˣ dx
- du = cos x dx, v = eˣ
∫eˣ sin x dx = eˣ sin x - ∫eˣ cos x dx
Second pass on ∫eˣ cos x dx:
- u = cos x, dv = eˣ dx
- du = -sin x dx, v = eˣ
∫eˣ cos x dx = eˣ cos x - ∫eˣ(-sin x) dx = eˣ cos x + ∫eˣ sin x dx
Substitute back:
∫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
Add ∫eˣ sin x dx to both sides:
2∫eˣ sin x dx = eˣ(sin x - cos x)
Final answer: (eˣ/2)(sin x - cos x) + C
Example 4: ∫ln x dx
No product here, but you can still use integration by parts. Write ln x = 1 · ln x.
Step 1: u = ln x, dv = 1 dx
Step 2: du = 1/x dx, v = x
Step 3: Apply the formula.
∫ln x dx = x·ln x - ∫x(1/x) dx = x·ln x - ∫1 dx
Final answer: x·ln x - x + C
Comparison: Integration Techniques
| Technique | Use When | Key Requirement |
|---|---|---|
| Substitution | Chain rule inside integral | Recognize f(g(x)) |
| Integration by Parts | Product of two different function types | LIATE rule for choosing u |
| Partial Fractions | Rational functions with denominators that factor | Proper fraction form |
| Trig Substitution | Square roots with squares | Recognize a² - x², x² + a², or x² - a² |
Common Mistakes
- Picking u wrong. This is the #1 error. If your new integral looks worse, swap your choices.
- Forgetting the minus sign. The formula is uv - ∫v du. The minus is mandatory.
- Dropping the constant of integration. Always add +C at the end.
- Not simplifying between steps. Clean up ∫v du before trying to solve it.
Practical Tips
If integration by parts produces the same integral you started with, you're probably doing it right. This happens with eˣ sin x and eˣ cos x. Just solve for the original integral algebraically at the end.
Sometimes you'll need to apply the technique three times. Work through each pass methodically and don't skip steps.
For polynomials times trig or exponentials, differentiate the polynomial completely. After enough passes, the polynomial disappears entirely.
Quick Reference
Remember the formula:
∫u dv = uv - ∫v du
Remember LIATE for choosing:
L → I → A → T → E
And remember: if the first choice fails, flip it. The formula works both ways.