Integration by parts is the Product Rule integrated. It handles products that substitution cannot.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The formula
The integral of u dv equals uv minus the integral of v du. The second integral should be easier than the first.
Choosing u
Pick u to be the part that simplifies when differentiated. A polynomial factor is usually the right choice.
A useful order
Logarithms, then inverse trigonometric, algebraic, trigonometric, exponential. The earlier in that list, the better a choice for u.
Repeating it
A polynomial of degree n usually needs n applications, each reducing the degree by one.
The circular case
For an exponential times a trigonometric function the integral returns to itself. Solving algebraically for it finishes the job.
The product rule, run backwards
∫u dv = uv − ∫v du. It comes from integrating the product rule and rearranging. The technique trades one integral for another, so it only helps when the new one is easier.
Choosing u
LIATE — logarithmic, inverse trig, algebraic, trigonometric, exponential — is a reliable ordering for which factor to call u. It is a rule of thumb, not a theorem, but it works in most exam cases.
Sometimes it must be applied twice
Integrals like ∫x²eˣ dx need two applications. Occasionally the original integral reappears, in which case it can be solved for algebraically — an elegant trick worth recognising.
With definite integrals
The uv term is evaluated at both limits and the remaining integral keeps them. Forgetting to evaluate the uv term at the limits, or dropping the limits from the second integral, are both common slips.
Step 2: Try It Yourself
Tap and try it out.
Step 3: Watch an Example
One step at a time.
Watch Sana Choose u
Sana integrates x times eˣ.
- Step 1
Substitution fails, because the derivative of one factor is not present as the other.
Step 4: Your Turn
Practice makes it stick.
The Choice
Problem 1 of 2
Integrating x times eˣ. Which should be u? 1 x, 2 eˣ.
The Repeats
Problem 2 of 2
Integrating x³ times eˣ. How many applications of the formula are needed?
Split the Product
1 of 8
Integrating x² times eˣ. How many applications?
2 of 8
Integrating x times sin x. Which should be u? 1 x, 2 sin x.
3 of 8
Integrating ln x. Which should be u? 1 ln x, 2 dx.
4 of 8
If u = x then du = k dx. What is k?
5 of 8
If dv = eˣ dx then v = k times eˣ. What is k?
6 of 8
Integrating x⁴ times eˣ. How many applications?
7 of 8
Put the integration by parts process in order.
- 1Set dv to the rest and find v and du.
- 2Apply the formula.
- 3Evaluate the new integral, repeating if needed.
- 4Choose u as the factor that simplifies when differentiated.
8 of 8
Integrating x times cos x. Which should be u? 1 x, 2 cos x.
Step 5: Quick Check
Show what you know.
Question 1 of 2
Integrating x⁵ times eˣ. How many applications of the formula?
Question 2 of 2
How should u be chosen?
What You Learned
- Integration by parts is the Product Rule integrated.
- The integral of u dv equals uv minus the integral of v du.
- Choose u to be the factor that simplifies when differentiated.