Calculus

Integration by Parts and the LIATE Rule

Integration by parts trades one integral for another. Everything about the technique comes down to making sure the trade is in your favour.

By the Gauth editorial teamHow this content is produced and checkedUpdated 10 September 2026

The idea

Every integration technique is a differentiation rule read backwards. Substitution is the chain rule reversed. Integration by parts is the product rule reversed, and you can derive it in three lines.

Start from the product rule and integrate both sides:

(uv)^{'} = u^{'} v + u v^{'} ⟹ uv = \int u^{'} v d x + \int u v^{'} d x

Now move one integral across:

\int u v^{'} d x = uv - \int u^{'} v d x

In differential notation, with $d v = v^{'} d x$ and $d u = u^{'} d x$ , that is the version you will write on paper:

\int u d v = uv - \int v d u

Notice what the formula does not do. It does not evaluate anything. It exchanges the integral you cannot do for a different integral, plus a term you get for free. The exchange is only useful if the new integral is easier, and it is entirely possible to choose badly and end up with something worse. That is the whole reason LIATE exists.

Take $\int x e^{x} d x$ . Choosing $u = x$ means $d u = d x$ , and the new integrand loses its $x$ entirely — you are left with $\int e^{x} d x$ , which is trivial. Choosing $u = e^{x}$ instead gives $v = x^{2} /2$ and a new integral of $\frac{1}{2} x^{2} e^{x}$ , which is strictly harder than what you started with. Same formula, opposite outcomes.

When you need it

Integration by parts is for a product of two functions of different kinds, where neither factor is the derivative of the other. That last clause is the important one, because it draws the line between parts and substitution.

$\int 2 x e^{x^{2}} d x$ looks like a product, but $2 x$ is the derivative of $x^{2}$ , so this is a substitution problem and parts would be a detour. $\int x e^{x} d x$ looks almost identical, but $x$ is not the derivative of anything in the exponential, so substitution has nothing to grab and parts is correct.

Reliable cues that parts is the technique:

A polynomial multiplied by $sin$ , $cos$ , or an exponential: $\int x^{2} cos x d x$ .
Anything multiplied by a logarithm: $\int x^{2} ln x d x$ .
A lone logarithm or inverse trig function: $\int ln x d x$ , $\int arctan x d x$ .
A trigonometric function multiplied by an exponential: $\int e^{x} sin x d x$ . Expect the circular case.

Once you have decided parts applies, the only remaining decision is which factor becomes $u$ . Work through it below.

Choosing u with LIATE

Look at what is being multiplied inside the integral. Which of these describes it?

The method

1. Rule out substitution first. Check whether one factor is, up to a constant, the derivative of the inside of the other. If it is, substitute. Parts on a substitution problem usually still works, but it takes three times as long.

2. Choose u by LIATE, then sanity-check dv. LIATE is a default, not a law. The two real requirements are that $u$ should get simpler when differentiated and that you must actually be able to integrate $d v$ . If LIATE hands you a $d v$ you cannot integrate, override it.

3. Write all four quantities down before substituting. A small block of $u$ , $d u$ , $d v$ , $v$ takes ten seconds and removes the chance of pairing the wrong pieces. When you integrate $d v$ to get $v$ , do not add a constant — any constant cancels out of the final result, so the simplest choice is the right one.

4. Substitute, and keep the minus sign attached to the whole integral. Write $uv$ , then a minus, then the entire new integral including its own $d x$ . If the new integral itself needs parts again, bracket it before you expand, because the minus sign has to distribute across everything it produces.

5. Recognise the three endings. The new integral is either immediately doable, or needs another round of parts, or is a multiple of the integral you started with. In the third case, stop: name the original integral $I$ , write the equation, and solve for $I$ the way you would solve any linear equation.

Worked examples

Worked example

\int x e^{x} d x

1
Algebraic beats exponential in LIATE, so u is the power of x. Build all four pieces.
$u = x, d u = d x d v = e^{x} d x, v = e^{x}$
2
Substitute into uv minus the integral of v du.
$\int x e^{x} d x = x e^{x} - \int e^{x} d x$
3
The remaining integral is immediate. This is the whole point of choosing u = x: differentiating it removed the x.
$= x e^{x} - e^{x} + C$
4
Factor, then check by differentiating: the product rule gives e^x(x-1) + e^x(1) = xe^x, which is the original integrand.
$= e^{x} (x - 1) + C$

Answer

\int x e^{x} d x = e^{x} (x - 1) + C

Worked example

\int x^{2} ln x d x

1
Logarithmic outranks algebraic in LIATE, so u is the logarithm even though the power looks like the friendlier choice.
$u = ln x, d u = \frac{d x}{x} d v = x^{2} d x, v = \frac{x ^{3}}{3}$
2
Substitute. The logarithm is now gone from the remaining integral, which is exactly the trade you wanted.
$= \frac{x ^{3}}{3} ln x - \int \frac{x ^{3}}{3} \cdot \frac{1}{x} d x$
3
Simplify the new integrand by cancelling one power of x.
$= \frac{x ^{3}}{3} ln x - \int \frac{x ^{2}}{3} d x$
4
Integrate the power. Note the denominator becomes 9, not 3, because the one-third is carried through the integration.
$= \frac{x ^{3}}{3} ln x - \frac{x ^{3}}{9} + C$

Answer

\int x^{2} ln x d x = \frac{x ^{3}}{3} ln x - \frac{x ^{3}}{9} + C

Worked example

\int e^{x} sin x d x

1
Neither factor simplifies when differentiated, so this will cycle. Name the integral I so you can refer to it later, and take u as the trigonometric factor.
$I = \int e^{x} sin x d x, u = sin x, d v = e^{x} d x$
2
First application. The new integral is no simpler, but it is no worse either.
$I = e^{x} sin x - \int e^{x} cos x d x$
3
Apply parts again to the new integral, keeping the same style of choice: trigonometric as u, exponential as dv. Switching now would undo the first step.
$\int e^{x} cos x d x = e^{x} cos x + \int e^{x} sin x d x = e^{x} cos x + I$
4
Substitute that back in. The original integral has reappeared, so this is now an equation in I rather than an integration problem.
$I = e^{x} sin x - e^{x} cos x - I$
5
Solve for I. Add I to both sides and halve. The constant of integration is added at the very end, after the division.
$2 I = e^{x} (sin x - cos x) ⟹ I = \frac{e ^{x} ( sin x - cos x )}{2} + C$

Answer

\int e^{x} sin x d x = \frac{e ^{x} ( s i n x - c o s x )}{2} + C

Try one of your own

Type an integral and you will get the choice of $u$ and $d v$ justified before the formula is applied, which is the part most solutions skip. The integral calculator with steps handles definite and indefinite versions and shows the boundary term evaluated separately. If you are not yet sure which technique a problem calls for, the guide to choosing an integration method walks every branch.

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Where people go wrong

Choosing u to make the first line look neat

The formula produces a boundary term and a new integral, and only the second one matters. Faced with $\int x ln x d x$ , people often pick $u = x$ because it differentiates to 1 — but then $d v = ln x d x$ , and finding $v$ means integrating $ln x$ , which is the harder problem you were avoiding. Judge a choice by the integral it leaves behind, not by how tidy the first line is.

Letting the minus sign apply to only part of what follows

When the second integral needs parts again, it produces two terms, and the minus sign in front of it applies to both. Bracketing the whole second application before expanding is the fix. For $\int x^{2} e^{x} d x$ the correct result is $e^{x} (x^{2} - 2 x + 2) + C$ ; a sign that fails to distribute typically gives $e^{x} (x^{2} - 2 x - 2)$ , and differentiating your answer catches it instantly.

Reversing the choice on the second round

In the circular case, the second application must keep the same kind of choice as the first. If you take $u = sin x$ and then switch to $u = e^{x}$ , the algebra collapses to $I = I$ — true, but useless. You have simply integrated by parts and then unintegrated by parts. Consistency in the choice is what makes the coefficient of $I$ come out as $- 1$ rather than $+ 1$ .

Applying the limits to only the remaining integral

For a definite integral the formula is $[uv]_{a}^{b} - \int_{a}^{b} v d u$ , and the boundary term needs evaluating too. Take $\int_{0}^{1} x e^{x} d x$ : the boundary term is $[x e^{x}]_{0}^{1} = e$ , the remaining integral is $[e^{x}]_{0}^{1} = e - 1$ , and the answer is $e - (e - 1) = 1$ . Forgetting to evaluate $uv$ leaves a stray function of $x$ in what should be a number, which is at least an error you can spot.

Practice

Include the constant of integration where the integral is indefinite. The last one is definite, so the answer is a single number.

Integration by parts practice

Answers are checked here — nothing is sent anywhere.

1
$\int x cos x d x$
2
$\int ln x d x$
3
$\int x e^{2 x} d x$
4
$\int_{0}^{1} x e^{x} d x$

Frequently asked questions

What does LIATE actually stand for?

Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Whichever of those categories appears earliest in that list becomes u; the rest of the integrand becomes dv. The ordering is not arbitrary — it ranks functions roughly by how much simpler they get when differentiated and how easily the remainder can be integrated.

Is LIATE a theorem or a rule of thumb?

A rule of thumb, and it has exceptions. It is a good default because it works on the large majority of textbook integrals, but the real criterion is the one it approximates: u should get simpler when you differentiate it, and dv must be something you can actually integrate. If LIATE gives you a dv you cannot integrate, override it.

What do I do when integration by parts goes in a circle?

If after two applications the original integral reappears with a coefficient, treat it as an unknown. Call the integral I, write the equation you have derived, and solve for I algebraically. That is how ∫e^x sin x dx is done. If instead the original reappears with coefficient exactly 1 and nothing else changed, you have undone your own first step — swap which factor you called u.

How do I integrate ln x when there is nothing to pair it with?

Let dv = dx. Then u = ln x, du = dx/x, and v = x, giving x ln x − ∫x · (1/x) dx = x ln x − x + C. The same trick handles arctan x, which becomes x arctan x − ½ ln(1+x²) + C. Whenever a single awkward function has no obvious partner, pairing it with dx = 1 · dx is the move.

How does integration by parts change for a definite integral?

The boundary term gets evaluated at the limits as well: ∫ from a to b of u dv equals [uv] from a to b, minus ∫ from a to b of v du. The commonest mistake is applying the limits only to the remaining integral and forgetting the uv piece. Evaluate both, and do not add a constant of integration.

Keep going

Integral calculator with stepsDefinite and indefinite integrals with the technique chosen and justified.U-substitutionThe technique to rule out before you commit to parts.The product and quotient rulesThe differentiation rule this whole technique is built from.Which integration technique should you use?A decision tree across every method, with the dead ends marked.