The Integration by Parts Formula
Integration by parts is derived from the product rule for differentiation and is used to integrate the product of two functions.
∫u dv = uv − ∫v du. In words: choose one part of the integrand to be u (which you will differentiate) and the other part to be dv (which you will integrate). Then compute du (the derivative of u) and v (the antiderivative of dv). Plug everything into the formula. The goal is to transform the original integral into a simpler integral ∫v du that you can evaluate.For example, to evaluate ∫x eˣ dx, let u = x and dv = eˣ dx.
Then du = dx and v = eˣ.
The formula gives xeˣ − ∫eˣ dx = xeˣ − eˣ + C.
Notice how the new integral ∫eˣ dx is much simpler than the original.
xeˣ − eˣ + CThe art of integration by parts lies entirely in choosing u and dv wisely so that the resulting integral is easier, not harder.
The LIATE Mnemonic for Choosing u
The most common question students ask is: how do I decide which part to call u and which to call dv? The LIATE rule provides a reliable guideline.
For ∫x cos(x) dx, x is algebraic (A) and cos(x) is trigonometric (T). Since A comes before T in LIATE, let u = x.
For ∫ln(x) dx, treat this as ∫ln(x) · 1 dx. Logarithmic (L) comes before algebraic (A), so u = ln(x) and dv = dx.
This gives x ln(x) − ∫x · (1/x) dx = x ln(x) − x + C.
x ln(x) − x + CLIATE is a heuristic, not an absolute rule, but it guides you to the correct choice in the vast majority of problems.
Step-by-Step Process
Here is a systematic approach to integration by parts that you can follow for any problem.
- Step 1
Identify the integrand as a product of two function types (e.g., algebraic times exponential).
- Step 2
Use LIATE to choose u (the part you differentiate) and dv (the rest, which you integrate).
- Step 3
Compute du by differentiating u, and compute v by integrating dv.
- Step 4
Substitute into the formula
∫u dv = uv − ∫v du. - Step 5
Evaluate the new integral
∫v du. If it is still complex, apply integration by parts again. - Step 6
Simplify and add the constant of integration C.
Let's apply this to ∫x² sin(x) dx.
Let u = x², dv = sin(x) dx. Then du = 2x dx, v = −cos(x).
Formula: −x² cos(x) + ∫2x cos(x) dx.
The new integral still needs integration by parts: let u = 2x, dv = cos(x) dx. Then du = 2 dx, v = sin(x).
Result: −x² cos(x) + 2x sin(x) − ∫2 sin(x) dx = −x² cos(x) + 2x sin(x) + 2cos(x) + C.
−x² cos(x) + 2x sin(x) + 2cos(x) + CThe Tabular Method for Repeated Integration by Parts
When you need to apply integration by parts multiple times — typically when one factor is a polynomial — the tabular method (also called the DI method) provides a fast shortcut.
- Set up two columns
Create two columns: one for successive derivatives of u and one for successive integrals of dv.
- Alternate the signs
Alternate the signs starting with positive.
- Multiply diagonally
Multiply diagonally and add up all the products.
For ∫x³ eˣ dx, list derivatives of x³ in the left column: x³, 3x², 6x, 6, 0.
List integrals of eˣ in the right column: eˣ, eˣ, eˣ, eˣ, eˣ.
Signs alternate: +, −, +, −.
Multiply diagonally: +x³ eˣ − 3x² eˣ + 6x eˣ − 6eˣ.
Factor: eˣ(x³ − 3x² + 6x − 6) + C.
eˣ(x³ − 3x² + 6x − 6) + CThe tabular method eliminates the need to set up integration by parts repeatedly, reducing the chance of sign errors and saving significant time. It works beautifully any time u is a polynomial, because the derivatives eventually reach zero and the table terminates naturally. Solver AI uses this technique internally and can show you the tabular layout step by step.
Common Pitfalls to Avoid
Even students who understand the formula often make avoidable errors.
If your choice makes the new integral harder than the original, stop and switch your assignments. For ∫eˣ sin(x) dx, either choice works, but you must apply integration by parts twice and solve the resulting equation for the integral.
The formula has a subtraction: uv − ∫v du. When applying integration by parts multiple times, sign errors accumulate quickly. Write each step explicitly.
For indefinite integrals, always include + C at the end.
If your integrand is a product of two different types of functions (like a polynomial times an exponential), integration by parts is likely the right tool. If it is a composition (like e^(x²)), try substitution instead.
For integrals like ∫eˣ cos(x) dx, after two rounds of integration by parts, you get the original integral on the right side. Solve the equation algebraically to find the answer rather than looping forever.
Worked Examples
∫ln(x) dx. Let u = ln(x), dv = dx.
Then du = (1/x) dx, v = x.
Result: x ln(x) − ∫1 dx = x ln(x) − x + C.
x ln(x) − x + C∫x² eˣ dx. Apply integration by parts twice (or use the tabular method).
eˣ(x² − 2x + 2) + C∫eˣ sin(x) dx. Let u = sin(x), dv = eˣ dx.
After two applications: ∫eˣ sin(x) dx = eˣ sin(x) − eˣ cos(x) − ∫eˣ sin(x) dx.
Add the integral to both sides: 2∫eˣ sin(x) dx = eˣ(sin(x) − cos(x)), so ∫eˣ sin(x) dx = eˣ(sin(x) − cos(x))/2 + C.
eˣ(sin(x) − cos(x))/2 + C∫arctan(x) dx. Let u = arctan(x), dv = dx.
Then du = 1/(1 + x²) dx, v = x.
Result: x arctan(x) − ∫x/(1 + x²) dx = x arctan(x) − ½ ln(1 + x²) + C.
x arctan(x) − ½ ln(1 + x²) + CPractice these types with Solver AI to build confidence and speed before your next exam.