The two most commonly tested integration techniques are u-substitution and integration by parts. Students frequently know both methods but not when to apply which. This guide gives the decision rule, explains the underlying logic, and covers the edge cases.
The decision rule
U-substitution works when the integrand contains a composite function and its derivative (or a scalar multiple of its derivative).
Integration by parts works when the integrand is a product of two different types of functions — typically from different categories: logarithms, inverse trig, polynomials, exponentials, trig functions.
The LIATE heuristic for integration by parts
When selecting u in integration by parts (∫u dv), choose u from whichever category appears first in this priority order:
L — Logarithmic functions (ln x, log x) I — Inverse trig functions (arcsin, arctan, etc.) A — Algebraic functions (polynomials like x², 3x+1) T — Trig functions (sin x, cos x) E — Exponential functions (e^x, 2^x)
The function later in the list becomes dv (and you integrate it to find v).
Why this heuristic works
Integration by parts formula: ∫u dv = uv − ∫v du
The goal is to choose u so that du is simpler than u, and dv so that v is manageable. Logarithms become fractions (simpler), inverse trig becomes algebraic (simpler), polynomials reduce degree when differentiated. Exponentials and trig functions cycle back to themselves — useful to keep as dv because integrating them is straightforward.
Worked example — substitution
∫ 2x · cos(x²) dx
The derivative of x² is 2x, and 2x appears in the integrand. Let u = x², du = 2x dx.
∫ cos(u) du = sin(u) + C = sin(x²) + C.
Worked example — integration by parts
∫ x · e^x dx
Two different function types: polynomial (x) and exponential (e^x). LIATE says let u = x (algebraic, higher priority) and dv = e^x dx.
du = dx, v = e^x
∫ x · e^x dx = x · e^x − ∫ e^x dx = x · e^x − e^x + C = e^x(x − 1) + C.
The case where you do both
∫ x · sin(x²) dx
There is a composition (x² inside sin) and a loose factor of x outside. Let u = x², du = 2x dx:
= ½ ∫ sin(u) du = −½ cos(u) + C = −½ cos(x²) + C.
This is substitution alone — the x in the integrand is the derivative of the inner function.
The case where you use parts twice
∫ e^x · sin(x) dx
Both e^x and sin x are in the "lower priority" LIATE categories. Apply parts twice (with the same choice of u each time) to create an equation that you solve for the integral:
Let u = sin x, dv = e^x dx: = e^x sin x − ∫ e^x cos x dx
Apply parts again to ∫ e^x cos x dx, let u = cos x, dv = e^x dx: = e^x cos x + ∫ e^x sin x dx
Now call the original integral I: I = e^x sin x − e^x cos x − I 2I = e^x(sin x − cos x) I = ½ e^x(sin x − cos x) + C.
Frequently Asked Questions
What if LIATE gives the wrong answer?
LIATE is a heuristic, not a law. There are integrals where the "wrong" choice of u still works — it just takes more steps. If LIATE gives a mess, try swapping your u and dv choices.
Is tabular integration the same as integration by parts?
Tabular integration is a systematic way to apply integration by parts multiple times when one function in the product is a polynomial. It produces the same answer, more efficiently, when the polynomial reduces to zero after several differentiations.