Integration by Parts Explained: Formula, Proof & Examples

Integration by parts is the product rule run backward: $\int u\,dv = uv - \int v\,du$. Use it to integrate a product when one factor becomes simpler after you differentiate it, as in $\int x\sin x\,dx$ or $\int \ln x\,dx$. This page gives the formula for indefinite and definite integrals, its three-line proof, the LIATE rule for choosing $u$, six worked examples and the tabular shortcut.

Contents11 sections

The formula at a glance

$$\int u\,dv = uv - \int v\,du$$

The same rule in function notation, and for a definite integral:

$$\int f(x)\,g'(x)\,dx = f(x)\,g(x) - \int f'(x)\,g(x)\,dx,\qquad \int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du.$$

1. Choose $u$ with the LIATE order; everything else (including $dx$) is $dv$.
2. Compute $du$ by differentiating and $v$ by integrating.
3. Substitute into $uv - \int v\,du$.
4. Finish the new integral, which should be simpler than the original.

What Is Integration by Parts? (The Definition)

Integration by parts is a rule for integrating a product of two functions. It does not compute the integral directly: it exchanges $\int u\,dv$ for $\int v\,du$. The trade pays off when differentiating $u$ makes it simpler (a polynomial loses a degree, $\ln x$ becomes $\frac1x$) while integrating $dv$ does not make it worse ($e^x$, $\sin x$ and $\cos x$ stay the same size).

The Integration by Parts Formula

$$\int u\,dv = uv - \int v\,du$$

Here $u$ is the factor you will differentiate and $dv$ is the factor you will integrate. Read it as “$u$ times $v$, minus the integral of $v\,du$“. You do not need a constant when you find $v$ from $dv$: any constant would cancel out of the final answer, so add $+C$ only once, at the end.

There is no separate rule called “differentiation by parts”. Its counterpart for derivatives is the product rule, $(uv)' = u'v + uv'$, and integration by parts is exactly that rule integrated, as the proof below shows.

Where Does the Formula Come From? (The Proof)

Derivation of the Integration by Parts Formula

Start from the product rule for two differentiable functions $u(x)$ and $v(x)$:

$$\frac{d}{dx}\big(uv\big) = u'v + uv'.$$

Integrate both sides with respect to $x$. The left side integrates back to $uv$:

$$uv = \int u'v\,dx + \int uv'\,dx.$$

Move one integral to the other side, and write $du = u'\,dx$, $dv = v'\,dx$:

$$\int u\,dv = uv - \int v\,du.$$

For a definite integral, integrate the product rule from $a$ to $b$ instead. The left side becomes $\big[uv\big]_a^b$, which gives the definite form above.

When to Use Integration by Parts: Conditions and the LIATE Rule

When Should You Use It

• The integrand is a product of two different kinds of function: $x e^x$, $x^2\cos x$, $e^x\sin x$.
• The integrand is a single function with no simple antiderivative but a simple derivative: $\ln x$, $\arctan x$, $\arcsin x$. Take $dv = dx$.
• Not when the integrand contains a function and its derivative, such as $x\,e^{x^2}$: that is a substitution.

The LIATE Rule: Choosing u Intelligently

Pick $u$ as the function that comes first in this list; the other factor goes into $dv$.

Letter	Function type	Examples
L	Logarithmic	$\ln x$, $\log_2 x$
I	Inverse trigonometric	$\arctan x$, $\arcsin x$
A	Algebraic (polynomials, powers)	$x$, $x^3$, $\sqrt{x}$
T	Trigonometric	$\sin x$, $\cos 2x$
E	Exponential	$e^x$, $2^x$

The logic: functions near the top get simpler when differentiated, functions near the bottom are easy to integrate. LIATE is a guideline, not a law. For $\int x^3 e^{x^2}dx$, the choice that works is $u = x^2$, $dv = x\,e^{x^2}dx$, because $x\,e^{x^2}$ is easy to integrate and $e^{x^2}$ alone is not.

Conditions for the Formula to Apply

The formula holds when $u$ and $v$ are differentiable with continuous derivatives on the interval of integration. For improper integrals, the boundary term $\big[uv\big]$ must have a finite limit; this is how $\int_0^\infty x e^{-x}dx = 1$ is computed.

Step-by-Step Examples of Integration by Parts

Example 1: A Polynomial Times a Trigonometric Function

Evaluate $\int x\sin x\,dx$.

LIATE puts the algebraic factor first: $u = x$, $dv = \sin x\,dx$. Then $du = dx$ and $v = -\cos x$.

$$\int x\sin x\,dx = -x\cos x - \int(-\cos x)\,dx = -x\cos x + \sin x + C.$$

The other choice, $u = \sin x$ and $dv = x\,dx$, leads to $\int \frac{x^2}{2}\cos x\,dx$: a harder integral than the one we started with.

Example 2: A Logarithm Using the “Multiply by 1” Trick

Evaluate $\int \ln x\,dx$.

There is only one factor, so write $\ln x = \ln x\cdot 1$: $u = \ln x$, $dv = dx$. Then $du = \frac{dx}{x}$ and $v = x$.

$$\int \ln x\,dx = x\ln x - \int x\cdot\frac1x\,dx = x\ln x - x + C.$$

Example 3: Cyclic Integration by Parts

Evaluate $I = \int e^x\cos x\,dx$.

Take $u = \cos x$, $dv = e^x dx$: $I = e^x\cos x + \int e^x\sin x\,dx$. Apply the rule again to the new integral with $u = \sin x$, $dv = e^x dx$:

$$I = e^x\cos x + e^x\sin x - \int e^x\cos x\,dx = e^x(\cos x + \sin x) - I.$$

The original integral came back. Solve for it: $2I = e^x(\sin x + \cos x)$, so

$$\int e^x\cos x\,dx = \frac{e^x(\sin x + \cos x)}{2} + C.$$

Keep the same type of function in $u$ both times (here, the trigonometric one). Switching to $u = e^x$ on the second step undoes the first step.

Example 4: Applying the Rule Twice

Evaluate $\int x^2 e^x\,dx$.

First, $u = x^2$, $dv = e^x dx$: $\int x^2e^x\,dx = x^2e^x - 2\int x e^x\,dx$. Then $u = x$, $dv = e^x dx$ gives $\int xe^x\,dx = xe^x - e^x$. Altogether:

$$\int x^2e^x\,dx = x^2e^x - 2\left(xe^x - e^x\right) + C = e^x\left(x^2 - 2x + 2\right) + C.$$

Each application lowers the power of $x$ by one. For higher powers, the tabular method below does all the steps at once.

Example 5: An Inverse Trigonometric Function

Evaluate $\int \arctan x\,dx$.

As in Example 2, $u = \arctan x$, $dv = dx$: $du = \frac{dx}{1+x^2}$, $v = x$. The remaining integral is a substitution with $w = 1 + x^2$:

$$\int \arctan x\,dx = x\arctan x - \int\frac{x}{1+x^2}\,dx = x\arctan x - \frac12\ln\left(1+x^2\right) + C.$$

Example 6: A Definite Integral

Evaluate $\int_0^1 x e^x\,dx$.

With $u = x$, $dv = e^x dx$, use the definite form and evaluate the boundary term right away:

$$\int_0^1 xe^x\,dx = \Big[xe^x\Big]_0^1 - \int_0^1 e^x\,dx = e - (e - 1) = 1.$$

The Tabular Method: Integration by Parts at Speed

How the Tabular Method Works

When $u$ is a polynomial, repeated integration by parts follows a fixed pattern. The tabular (or DI) method writes it as a table: differentiate $u$ down one column until you reach $0$, integrate $dv$ down the other, and attach alternating signs $+, -, +, -$.

Sign	D (differentiate)	I (integrate)
$+$	$x^3$	$\cos x$
$-$	$3x^2$	$\sin x$
$+$	$6x$	$-\cos x$
$-$	$6$	$-\sin x$
	$0$	$\cos x$

Multiply each D entry by the I entry one row below and apply its sign:

$$\int x^3\cos x\,dx = x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + C.$$

Graphical Interpretation of Integration by Parts

Plot the curve traced by the point $(u, v)$ as $x$ goes from $a$ to $b$. The area between the curve and the $v$-axis is $\int u\,dv$, and the area between the curve and the $u$-axis is $\int v\,du$. Together they fill the difference of two rectangles, $u(b)v(b) - u(a)v(a)$. That is the definite formula: $\int u\,dv + \int v\,du = \big[uv\big]_a^b$.

Key Properties and Special Cases

Reduction Formulas

Applying the rule once to an integral with a power $n$ often gives the same integral with power $n-1$ or $n-2$. That relation is a reduction formula:

$$\int x^n e^x\,dx = x^n e^x - n\int x^{n-1}e^x\,dx,\qquad \int \sin^n x\,dx = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}\int\sin^{n-2}x\,dx.$$

Repeating the first one until $n = 0$ gives $\int x^n e^x\,dx$ for any $n$; the second one handles every power of $\sin x$.

Integration by Parts for Definite Integrals

$$\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du$$

Evaluate the boundary term $\big[uv\big]_a^b$ as soon as it appears; there is no need to find the full antiderivative first. Example: $\int_0^{\pi/2} x\cos x\,dx = \big[x\sin x\big]_0^{\pi/2} - \int_0^{\pi/2}\sin x\,dx = \frac{\pi}{2} - 1$.

Common Mistakes

• Losing the minus sign in $uv - \int v\,du$, especially when $v$ itself is negative ($v = -\cos x$).
• Choosing $u$ backward. If the new integral is harder than the original, swap the roles of $u$ and $dv$.
• Using it where substitution is needed. $\int x\cos(x^2)\,dx$ is a substitution ($w = x^2$), not a product to split.
• Forgetting the boundary term in a definite integral, or evaluating it at only one limit.

Practice Problems

Ready to try it yourself? The integration by parts practice problems page has problems from easy to hard, each with a hint and a full step-by-step solution.

Frequently asked questions