What Is Integration by Parts?
Integration by parts is a method for integrating the product of two functions. When an integrand is one function multiplied by another, and neither simple substitution nor a standard rule cracks it, this technique rewrites the integral into a form you can finish. The formula is:
$$\int u,dv = uv - \int v,du$$
Here you split the integrand into two pieces: a part you call $u$ (which you will differentiate) and a part you call $dv$ (which you will integrate). From those you build $du$ by differentiating $u$, and $v$ by integrating $dv$. The formula then trades the original integral $\int u,dv$ for the usually easier $\int v,du$.
The definite-integral version simply carries the limits through:
$$\int_a^b u,dv = \Big[uv\Big]_a^b - \int_a^b v,du$$
The technique is the integration counterpart of the product rule for derivatives, and it sits alongside substitution as one of the two workhorse methods of integration. A published reference for the same formula lives at the Bhanzu integration of uv page.
How Do You Derive The Integration by Parts Formula?
The formula is not a guess. It is the product rule read backwards.
Start from the product rule for the derivative of a product $u(x),v(x)$:
$$\frac{d}{dx}\big[u,v\big] = u,\frac{dv}{dx} + v,\frac{du}{dx}$$
Integrate both sides with respect to $x$. The left side integrates back to $uv$:
$$u,v = \int u,\frac{dv}{dx},dx + \int v,\frac{du}{dx},dx$$
Rewrite the two integrals in differential form as $\int u,dv$ and $\int v,du$:
$$u,v = \int u,dv + \int v,du$$
Now solve for the integral you want by moving the other term across:
$$\int u,dv = uv - \int v,du$$
That single rearrangement is the whole method. Because it inverts the product rule, integration by parts is exactly the right tool whenever the integrand looks like something that came from differentiating a product.
What Is The LIATE Rule For Choosing u?
The one decision that makes or breaks the method is which factor to name $u$. A good choice for $u$ gets simpler when differentiated; the leftover $dv$ should be something you can integrate without trouble. The LIATE rule ranks the function types in the order you should prefer for $u$:
Table: The LIATE priority order for choosing $u$ (first match on the list becomes $u$).
Priority | Letter | Function type | Examples |
|---|---|---|---|
1st | L | Logarithmic | $\ln x$, $\log x$ |
2nd | I | Inverse trigonometric | $\arctan x$, $\arcsin x$ |
3rd | A | Algebraic (powers) | $x$, $x^2$, $3x+1$ |
4th | T | Trigonometric | $\sin x$, $\cos x$ |
5th | E | Exponential | $e^x$, $2^x$ |
Whichever factor appears earlier on the list becomes $u$; the other becomes $dv$. Some books call the same idea ILATE, swapping the first two letters. The order works because logarithms and inverse trig functions get much simpler when differentiated (so you want them as $u$), while exponentials barely change when integrated (so they are safe as $dv$). LIATE is a strong default, not a law, and the FAQ below shows a case where you deliberately break it.
How Do You Use Integration by Parts? (Worked Examples)
Each example follows the same four moves: pick $u$ and $dv$ by LIATE, build $du$ and $v$, substitute into $\int u,dv = uv - \int v,du$, then differentiate the answer back to confirm it.
Example 1: Evaluate $\int x e^x,dx$.
This is Algebraic times Exponential. LIATE puts A before E, so $u = x$ and $dv = e^x,dx$.
$$u = x ;\Rightarrow; du = dx, \qquad dv = e^x,dx ;\Rightarrow; v = e^x$$
Substitute into the formula:
$$\int x e^x,dx = x e^x - \int e^x,dx = x e^x - e^x + C$$
Check by differentiating the result: $\frac{d}{dx}\big[x e^x - e^x\big] = \big(e^x + x e^x\big) - e^x = x e^x$, which is the original integrand.
Final answer: $\int x e^x,dx = x e^x - e^x + C$.
Example 2: Evaluate $\int x \cos x,dx$.
Algebraic times Trigonometric. A comes before T, so $u = x$ and $dv = \cos x,dx$.
$$u = x ;\Rightarrow; du = dx, \qquad dv = \cos x,dx ;\Rightarrow; v = \sin x$$
$$\int x \cos x,dx = x \sin x - \int \sin x,dx = x \sin x + \cos x + C$$
Check: $\frac{d}{dx}\big[x \sin x + \cos x\big] = \big(\sin x + x \cos x\big) - \sin x = x \cos x$. Correct.
Final answer: $\int x \cos x,dx = x \sin x + \cos x + C$.
Example 3: Evaluate $\int \ln x,dx$.
There is only one visible factor, but writing $\int \ln x,dx = \int (\ln x)(1),dx$ turns it into a product. LIATE ranks Logarithmic first, so $u = \ln x$ and $dv = 1,dx$.
$$u = \ln x ;\Rightarrow; du = \frac{1}{x},dx, \qquad dv = dx ;\Rightarrow; v = x$$
$$\int \ln x,dx = x \ln x - \int x \cdot \frac{1}{x},dx = x \ln x - \int 1,dx = x \ln x - x + C$$
Check: $\frac{d}{dx}\big[x \ln x - x\big] = \Big(\ln x + x \cdot \tfrac{1}{x}\Big) - 1 = \ln x + 1 - 1 = \ln x$. Correct. This trick, pairing a lone function with $dv = dx$, also integrates $\arctan x$ and other logarithmic functions.
Final answer: $\int \ln x,dx = x \ln x - x + C$.
Example 4 (definite integral): Evaluate $\int_0^1 x e^x,dx$.
Use the same antiderivative found in Example 1, then apply the limits:
$$\int_0^1 x e^x,dx = \Big[x e^x - e^x\Big]_0^1$$
$$= \big(1 \cdot e^1 - e^1\big) - \big(0 \cdot e^0 - e^0\big) = (e - e) - (0 - 1) = 0 - (-1) = 1$$
Final answer: $\int_0^1 x e^x,dx = 1$.
What Is The Tabular (Repeated) Method?
When $u$ is a power of $x$ that will differentiate down to zero and $dv$ is easy to integrate again and again (like $e^x$, $\sin x$, or $\cos x$), applying the formula repeatedly is tedious. The tabular shortcut organises the repetition.
Take $\int x^2 e^x,dx$. List the derivatives of $x^2$ in one column until you reach $0$, and the successive integrals of $e^x$ in the next, then combine along the diagonals with alternating signs $+, -, +$:
Table: Tabular integration by parts for $\int x^2 e^x,dx$.
Sign | Derivatives of $x^2$ | Integrals of $e^x$ |
|---|---|---|
$+$ | $x^2$ | $e^x$ |
$-$ | $2x$ | $e^x$ |
$+$ | $2$ | $e^x$ |
$0$ | $e^x$ |
Multiply each derivative by the integral one row below it, attach the sign, and add:
$$\int x^2 e^x,dx = x^2 e^x - 2x e^x + 2 e^x + C = e^x\big(x^2 - 2x + 2\big) + C$$
Check: $\frac{d}{dx}\big[e^x(x^2 - 2x + 2)\big] = e^x(x^2 - 2x + 2) + e^x(2x - 2) = e^x \cdot x^2 = x^2 e^x$. Correct.
How Do You Handle The Circular Case?
Some integrals send you in a loop: apply parts twice and the original integral reappears. This circular pattern is not a dead end, it lets you solve for the integral with algebra. The classic case is $\int e^x \sin x,dx$.
Let $I = \int e^x \sin x,dx$. Choose $u = \sin x$, $dv = e^x,dx$, so $du = \cos x,dx$ and $v = e^x$:
$$I = e^x \sin x - \int e^x \cos x,dx$$
Apply parts again to the new integral, with $u = \cos x$, $dv = e^x,dx$, so $du = -\sin x,dx$ and $v = e^x$:
$$\int e^x \cos x,dx = e^x \cos x - \int e^x(-\sin x),dx = e^x \cos x + \int e^x \sin x,dx$$
Substitute back, noticing that $\int e^x \sin x,dx$ is just $I$ again:
$$I = e^x \sin x - \Big(e^x \cos x + I\Big) = e^x \sin x - e^x \cos x - I$$
Now solve for $I$ by adding $I$ to both sides, giving $2I = e^x \sin x - e^x \cos x$:
$$I = \frac{1}{2}e^x\big(\sin x - \cos x\big) + C$$
Check: $\frac{d}{dx}\Big[\tfrac{1}{2}e^x(\sin x - \cos x)\Big] = \tfrac{1}{2}\big[e^x(\sin x - \cos x) + e^x(\cos x + \sin x)\big] = \tfrac{1}{2}e^x(2\sin x) = e^x \sin x$. Correct. Working with $\sin x$ and $\cos x$ here leans on the differentiation of trigonometric functions.
Why Does Integration by Parts Work?
The algebra says the formula is the product rule reversed. The picture says something you can almost see.
The area interpretation. Plot $u$ against $v$. The product $uv$ is the area of the rectangle with sides $u$ and $v$. That rectangle is split into two curved regions by the graph relating the variables: one region has area $\int u,dv$ and the other has area $\int v,du$. The two areas add up to the whole rectangle $uv$, which rearranges into $\int u,dv = uv - \int v,du$.
Trading one integral for another. The method never makes an integral vanish. It swaps a hard integral for the term $uv$ (which needs no integration) plus a new integral $\int v,du$. The choice of $u$ and $dv$ is entirely about making that new integral simpler than the one you started with.
Why LIATE points the right way. Differentiating a logarithm or an inverse trig function replaces it with a plain algebraic fraction, so putting it in the $u$ slot melts away the awkward part. An exponential integrates to itself, so it is harmless in the $dv$ slot. LIATE is just this reasoning turned into an ordering.
Because the integral remains an area under a curve throughout, integration by parts is best read as a rule for reorganising an area into a rectangle minus a leftover, not as a trick with no meaning behind it.
Who Discovered Integration by Parts?
The method is older than the notation students use for it, and it grew directly out of the product rule that Leibniz wrote down when calculus was brand new.
Two names anchor the story:
Gottfried Wilhelm Leibniz (1646–1716, Germany) gave the product rule and the integral sign $\int$ that the derivation relies on.
Brook Taylor (1685–1731, England) first published the integration by parts formula in 1715.
Where Is Integration by Parts Used In The Real World?
The formula is not confined to textbooks. It appears wherever a product of two changing quantities has to be accumulated.
Physics and engineering: computing the work done by a variable force over a distance, and deriving the equations of motion in mechanics, routinely needs integration by parts.
Signal processing: the Fourier and Laplace transforms that filter audio and images are integrals of a signal multiplied by a wave, and integration by parts is a core step in evaluating and simplifying them.
Probability and statistics: the expected value of a continuous random variable is an integral of a value times a density, and parts converts it into a cleaner form (the tail-integral formula).
Differential equations: solving the equations that model heat flow, vibrations, and circuits often hinges on an integration-by-parts step, and it is the engine behind the "weak formulations" used in engineering simulation.
Numerical analysis: error estimates for approximation methods are frequently derived by applying the formula to bound a leftover integral.
One compact identity, $\int u,dv = uv - \int v,du$, quietly supports mechanics, communication systems, statistics, and simulation software. The same tool a student meets on paper is the one that clears the integrals behind those fields.
What Are The Most Common Mistakes With Integration by Parts?
These four errors account for most lost marks, verified against real student questions on Physics Forums (the "+C" confusion), Cuemath's own FAQ on when the rule fails, and standard graded calculus worksheets.
Choosing $u$ and $dv$ the wrong way round.
Where it slips in:
A student sets $u = e^x$ and $dv = x,dx$ in $\int x e^x,dx$, then finds the new integral is harder than the original.
Don't do this:
Do not pick $u$ at random. If the leftover integral $\int v,du$ looks worse than what you began with, the choice was backwards.
The correct way:
Use LIATE. Whichever factor comes first on Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential becomes $u$. For $\int x e^x,dx$, Algebraic beats Exponential, so $u = x$, not $e^x$.
Dropping the minus sign in $-\int v,du$.
Where it slips in:
A student writes $\int u,dv = uv + \int v,du$, or loses the sign on the second pass of a circular problem, and the answer fails to check.
Don't do this:
Do not misremember the formula. The sign is subtraction: $\int u,dv = uv - \int v,du$.
The correct way:
Write the formula out every time, and always differentiate your final answer back to the integrand. If the derivative does not reproduce the integrand, a sign is usually the culprit.
Forgetting $+C$ on the final indefinite integral.
Where it slips in:
A student adds no constant of integration at the end, having also worried needlessly about a separate constant when finding $v$ from $dv$.
Don't do this:
Do not omit $+C$ from an indefinite integral, and do not carry an extra constant when integrating $dv$ to get $v$ (that constant cancels and the simplest choice, taking it as zero, is correct).
The correct way:
Take the plainest $v$ (no added constant), work through the formula, and attach a single $+C$ to the final indefinite answer.
Mishandling the limits in a definite integral.
Where it slips in:
A student evaluates only the leftover integral at the bounds and forgets to evaluate the $uv$ term at both limits, or plugs the limits into $v$ too early.
Don't do this:
Do not apply the bounds to only part of the expression. Every piece of $\Big[uv\Big]_a^b - \int_a^b v,du$ is evaluated between $a$ and $b$.
The correct way:
Keep the whole result symbolic, then evaluate the completed antiderivative $uv - \int v,du$ at the upper limit minus the lower limit, exactly as in Example 4.
Practice Problems On Integration by Parts
Work each with $\int u,dv = uv - \int v,du$, then differentiate your answer to check. Answers follow each line.
Find $\int x \sin x,dx$.
(Answer: $-x \cos x + \sin x + C$.)Find $\int x e^{2x},dx$.
(Answer: $\tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x} + C$.)Find $\int x^2 \ln x,dx$.
(Answer: $\tfrac{x^3}{3}\ln x - \tfrac{x^3}{9} + C$.)Find $\int \arctan x,dx$.
(Answer: $x \arctan x - \tfrac{1}{2}\ln(1 + x^2) + C$.)Find $\int x \cos 2x,dx$.
(Answer: $\tfrac{1}{2}x \sin 2x + \tfrac{1}{4}\cos 2x + C$.)Evaluate $\int_1^e \ln x,dx$.
(Answer: using $x \ln x - x$, this is $(e \cdot 1 - e) - (1 \cdot 0 - 1) = 1$.)
Where Should You Go Next After Integration by Parts?
Integration by parts is one pillar of integral calculus, and a few natural doors open from here.
Integration of uv. The formula reference for parts, handy as a quick lookup while you practise.
Product rule (uv differentiation). Master the differentiation rule that integration by parts reverses, so the derivation feels obvious.
AP Calculus AB vs BC. See where integration by parts sits in the exam syllabus and which course covers it.
If your child is building these calculus foundations, a live Bhanzu trainer teaches integration by parts starting from the "why" (the product rule it reverses and the area it reorganises) in the Bhanzu math program.
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