Key takeaways
- Formula: ∫u dv = u·v − ∫v du.
- Choose u with LIATE: Logs, Inverse trig, Algebraic, Trig, Exponentials.
- ∫x·eˣ dx = eˣ(x − 1) + C.
- The #1 mistake is choosing u badly, which makes the new integral harder.
In this guide
What is integration by parts?
An integral adds up area under a curve, and finding one usually means reversing a derivative. That is easy for single functions, but hard when two types are multiplied, such as a power of x and an exponential.
Integration by parts comes straight from the product rule: (uv)′ = u′v + uv′. Integrate both sides and rearrange, and you get the formula.
Why does integration by parts matter?
Many important formulas contain a product like t·e^(−t) or t·sin t: in electronics, springs, probability and physics. Integration by parts is the standard way to handle them.
Integration by parts is the product rule running backwards, just as u-substitution is the chain rule running backwards. Those two methods solve most integrals you’ll meet in a first course.
The formula
| Symbol | What it means | Example (∫x·eˣ dx) |
|---|---|---|
| u | The part you differentiate. Choose something that gets simpler | x |
| du | The derivative of u, times dx | dx |
| dv | The part you integrate | eˣ dx |
| v | The integral of dv | eˣ |
To choose u, use LIATE: Logarithms, Inverse trig, Algebraic (powers of x), Trig, Exponentials. Pick u from whichever type comes first in that list.
This calculator finds the definite integral of xⁿ·eˣ from a to b, for whole numbers n from 0 to 5, applying the method n times.
How to integrate by parts step by step
- 1Choose u and dv. For ∫x eˣ dx: u = x, dv = eˣ dx.
- 2Find du and v. du = dx, v = eˣ.
- 3Plug into the formula.
x eˣ − ∫eˣ dx. - 4Finish the easier integral.
x eˣ − eˣ + C = eˣ(x − 1) + C. - 5For a definite integral, use the limits. Put in b and a, and subtract.
Now try it yourself. Enter the power n and the limits in the calculator below to see the result with its steps.
Try it yourself
Pre-filled with the example — change any value.
Common mistakes to avoid
- Sign errors. Keep the minus sign in front of ∫v du, especially when you repeat the method.
- Forgetting to repeat. For x²eˣ you need two rounds, one for each power of x.
- Dropping + C on an indefinite integral.
Tips and tricks
- Repeated parts:
∫x²eˣ dx = eˣ(x² − 2x + 2) + C. From 0 to 1 that givese − 2 ≈ 0.7183. - Hidden dv = dx: for
∫ln x dx, take u = ln x and dv = dx. The answer isx ln x − x + C. - Other methods: if one part is the derivative of something inside another part, try u-substitution first.
Where you’ll use it in real life
- Electronics: charging and discharging circuits often involve t·e^(−t).
- Mechanics: damped springs, like car suspension, involve products of exponentials and waves.
- Probability: the average waiting time for random events uses ∫t·e^(−t) dt.
- Signals: Fourier and Laplace methods rely on integration by parts.
Quick summary
Integration by parts turns ∫u dv into u·v − ∫v du. Choose u with LIATE so that the new integral is easier, then check by differentiating. Use the calculator above for ∫xⁿ·eˣ dx between any two limits, and see the derivative of eˣ for why eˣ is so easy to integrate.
Worked example
A worksheet asks you to find ∫ x·eˣ dx from x = 0 to x = 2 using integration by parts, with u = x and dv = eˣ dx. Here the power of x is n = 1, the lower limit is a = 0 and the upper limit is b = 2.
- 01∫[0 to 2] x^1 · eˣ dx
- 02Integration by Parts: u = x^1, dv = eˣ dx
- 03du = 1x^0dx, v = eˣ
- 04= [x^1·eˣ]|[0,2] − ∫[0,2] 1x^0·eˣ dx
- 05Repeat for each remaining power of x…
- 06F(2) = 7.3891
- 07F(0) = -1
- 08Result = F(2) − F(0) = 8.3891
These numbers are pre-filled in the calculator above.
Frequently asked questions
How do I choose u in integration by parts?
Use LIATE: pick u from the first type in the list Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. The rest becomes dv.
What is the integral of x times e^x?
It is x eˣ − eˣ + C, which you can also write as eˣ(x − 1) + C.
When should I use integration by parts instead of substitution?
Use substitution when one part is the derivative of something inside another part. Use integration by parts when two unrelated types are multiplied, like x and eˣ.
