§ 01Where the Formula Comes From
Integration by parts is not a trick, it is the product rule of differentiation run in reverse. Understanding that origin makes the formula memorable and tells you exactly when to use it.
Recall the product rule: if u and v are both differentiable functions of x, then the derivative of their product is
Rearranging gives us u dv/dx = d(uv)/dx − v du/dx. Now integrate both sides with respect to x:
That is the entire formula. The left-hand side is the integral you want to evaluate. The right-hand side trades it for a new integral, ∫ v du. The art of integration by parts is choosing u and dv so that the new integral is simpler than the one you started with. A poor choice makes the new integral harder, or produces an endless loop, experience and the LIATE rule (§ 03) prevent both.
A constant of integration C is only needed once, at the very end of the calculation, not at every intermediate step. Adding it too early creates clutter without adding any information, since a constant plus a constant is still just a constant.
§ 02Applying the Formula — Step by Step
Every integration-by-parts calculation follows the same mechanical procedure. Before you look at any examples, internalise these four steps so that applying the technique becomes automatic.
The Four Steps
- Step 1. Identify u and dv from the integrand (use LIATE).
- Step 2. Differentiate u to get du; integrate dv to get v.
- Step 3. Substitute into uv − ∫ v du.
- Step 4. Evaluate the remaining integral; add C.
What to Watch Out For
- When finding v from dv, pick the simplest antiderivative (set any constant to zero).
- If ∫ v du is still hard, you may need a second round of IBP.
- If the original integral reappears on the right, collect it algebraically.
- Always verify your answer by differentiating: it must return the original integrand.
Example 1 — A Polynomial Times an Exponential
Choose u = x and dv = ex dx. The polynomial goes to u because differentiating it reduces its degree. The exponential goes to dv because its integral is equally simple.
Example 2 — A Polynomial Times a Logarithm
Here the LIATE rule is critical. Set u = ln x (Logarithmic) and dv = x dx (Algebraic). If you swap them, you must integrate ln x directly: which requires IBP again, creating a circular problem.
§ 03The LIATE Rule — Choosing u Correctly
The hardest part of integration by parts is deciding which factor to call u and which to call dv. A wrong choice does not make the problem impossible, it just makes the resulting integral harder than the one you started with. The LIATE mnemonic gives a reliable priority order for u:
| Letter | Function Type | Why It Goes to u | Examples |
|---|---|---|---|
| L | Logarithmic | Differentiating removes the log; integrating a log requires IBP. Always put logs in u. | ln x, log2 x, ln(x² + 1) |
| I | Inverse Trig | Their derivatives are algebraic; their integrals are messy. Put them in u. | arcsin x, arctan x, arcsec x |
| A | Algebraic | Polynomials differentiate to lower-degree polynomials, simplifying ∫ v du. | x³, 5x² − 2, √x, x² + 3x |
| T | Trigonometric | Trig functions integrate and differentiate to other trig functions — they cycle. | sin x, cos x, tan x, sec x |
| E | Exponential | eax integrates easily and stays the same form. It is the best choice for dv. | ex, e3x, 2x, e−x |
To use LIATE: scan the integrand, identify the two factors, then assign u to whichever function appears earlier in the list (L before I before A before T before E). The remaining factor becomes dv. This heuristic works in the overwhelming majority of cases and is worth memorising before you encounter your first examination question on the technique.
Quick LIATE Assignments for Common Integrals
§ 04More Fully Worked Examples
The best way to absorb integration by parts is to work through a wide variety of problems. The examples below are ordered from straightforward to technically demanding.
Example 3 — Integrating ln x Alone
There is only one visible factor, but we can write 1 ⋅ ln x. Set u = ln x and dv = 1 dx, i.e. dv = dx. This is the "hidden" IBP.
Example 4 — Inverse Trig Function
Again write as 1 ⋅ arctan x. Set u = arctan x (Inverse trig, first in LIATE) and dv = dx.
For the remaining integral, let w = 1 + x², dw = 2x dx:
∫ x/(1 + x2) dx = (1/2) ln|1 + x2| = (1/2) ln(1 + x2)Example 5 — Applying IBP Twice
Set u = x2, dv = ex dx. Then du = 2x dx, v = ex.
∫ x2 ex dx = x2 ex − ∫ 2x ex dxThe new integral ∫ 2x ex dx requires another IBP. Set u = 2x, dv = ex dx. Then du = 2 dx, v = ex.
∫ 2x ex dx = 2x ex − ∫ 2ex dx = 2x ex − 2ex§ 05The Tabular Method (Column Integration)
When integration by parts must be applied multiple times, typically because one factor is a polynomial of degree 2 or higher, writing out the full formula at every step is slow and error-prone. The tabular method, also known as column integration or the DI method, condenses all those steps into a single table. It is not a different formula; it is simply a bookkeeping shortcut that keeps track of alternating signs automatically.
To construct the table, create three columns: Signs, D (the function you differentiate, which is your u choice), and I (the function you integrate, which is your dv choice). Alternate + and − signs down the first column, differentiate the D column repeatedly until it reaches zero, and integrate the I column the same number of times. Multiply diagonally and sum.
Example 6 — Tabular Method for ∫ x3 e2x dx
| Sign | D (differentiate) | I (integrate) |
|---|---|---|
| + | x3 | e2x |
| − | 3x2 | (1/2) e2x |
| + | 6x | (1/4) e2x |
| − | 6 | (1/8) e2x |
| + | 0 — stop | (1/16) e2x |
Read diagonally, multiplying each D entry by the I entry one row below it, with the sign in the same row as the D entry:
+ (−1) ⋅ 3x2 ⋅ (1/4)e2x
+ (+1) ⋅ 6x ⋅ (1/8)e2x
+ (−1) ⋅ 6 ⋅ (1/16)e2x
Verify by differentiating, that is, apply the product rule to each term in the bracket multiplied by e2x, and you will recover x3 e2x precisely. This verification step is not optional: it is the fastest way to catch arithmetic errors in a tabular calculation.
§ 06Cyclic Integrals — When the Original Integral Returns
Some integrands, notably eax sin bx and eax cos bx, have the curious property that after two applications of integration by parts, the original integral reappears on the right-hand side. This looks like a disaster (as if nothing has been achieved) but it is actually the key to solving the problem. Once the original integral appears on both sides, treat it as an unknown and solve algebraically.
Example 7 — ∫ ex sin x dx
Set u = sin x, dv = ex dx. Then du = cos x dx, v = ex.
∫ ex sin x dx = ex sin x − ∫ ex cos x dx …(I)Apply IBP to ∫ ex cos x dx. Set u = cos x, dv = ex dx. Then du = −sin x dx, v = ex.
∫ ex cos x dx = ex cos x + ∫ ex sin x dx …(II)Substitute (II) into (I). Let I stand for ∫ ex sin x dx:
I = ex sin x − [ex cos x + I] I = ex sin x − ex cos x − I 2I = ex(sin x − cos x)§ 07Definite Integrals by Parts
Integration by parts works for definite integrals with only a minor change: the boundary-evaluated term [uv] is computed at the limits, just like any antiderivative in the Fundamental Theorem of Calculus.
The term [uv]ab means u(b)v(b) − u(a)v(a). The remaining integral ∫ab v du is evaluated over the same interval. All other aspects of the technique, choosing u and dv, applying LIATE, handling cyclic integrals, remain identical to the indefinite case. The constant C disappears, because it cancels when you apply the limits.
Example 8 — Definite IBP
u = x, dv = ex dx → du = dx, v = ex.
§ 08Reduction Formulae
A reduction formula is a recurrence relation that expresses ∫ fn(x) dx in terms of ∫ fn−2(x) dx (or a similar lower-power integral). They are derived by a single application of integration by parts, followed by algebraic manipulation. Once the formula is established, integrals of arbitrarily high powers become mechanical.
Reduction Formula for ∫ sinn x dx
Derived by setting u = sinn−1 x, dv = sin x dx and applying IBP once.
Starting from this formula, you can compute ∫ sin4 x dx by applying it twice, ending at ∫ sin0 x dx = ∫ 1 dx = x. Similarly, ∫ sin5 x dx ends at ∫ sin x dx = −cos x. The same approach gives reduction formulae for cosn x, tann x, secn x, and even (ln x)n.
Example 9 — Using the Reduction Formula for ∫ sin4 x dx
§ 09Common Mistakes and How to Avoid Them
Most errors in integration by parts fall into one of five categories. Each is entirely preventable once you know what to look for.
§ 10Summary and Decision Guide
Integration by parts is the technique of choice whenever the integrand is a product of two functions from different categories: and specifically when substitution offers no clear path forward. Learning to recognise these integrands on sight and to execute the procedure quickly and accurately is one of the most valuable skills in single-variable calculus.
Can you substitute to simplify? → Try substitution first (it is faster)
Does LIATE give a clear u? → Apply the formula once
After one round, is the new integral simpler? → Continue to solution
Does the original integral reappear? → Collect it algebraically (cyclic case)
Is the polynomial degree 3 or higher? → Use the tabular method
The core formula ∫ u dv = uv − ∫ v du contains everything you need. Every variation (tabular integration, reduction formulae, cyclic collection) is just an efficient way of applying the same formula under particular structural conditions. There is no separate technique to memorise for each type: only a single underlying identity, and judgment about when and how to apply it.