Integration by Parts: Formula, Liate Rule and Examples

Integration by parts is a calculus technique. You use it to integrate the product of two functions. Its formula is $$ \int u \, dv = uv - \int v \, du $$. You select \(u\) according to the LIATE priority rule. This rule lists Logarithmic, Inverse trig, Algebraic, Trigonometric, and Exponential functions. You set \(dv\) as the remaining term. As a result, this transforms a difficult integral into a simpler one.

This technique turns an intractable integral into one you can solve. It works by differentiating one function and integrating the other. Therefore, it reverses the product rule for differentiation. It provides a structured approach. Use it when direct integration rules or substitution fail on product terms.

What is the integration by parts formula?

The standard indefinite integration by parts formula is written as:

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

For definite integrals evaluated between lower limit \(a\) and upper limit \(b\), the formula incorporates boundary evaluations:

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

In both formulas, \(u\) and \(v\) represent differentiable functions of \(x\). Also, \(du\) and \(dv\) represent their respective differentials. So, \(du = u'(x)dx\) and \(dv = v'(x)dx\). The formula’s core purpose is to trade a challenging integral, \(\int u \, dv\). You can trade it for a simpler integral, \(\int v \, du\).

Where does the integration by parts formula come from?

The formula comes directly from the product rule of differentiation. First, begin with two differentiable functions. These are \(u(x)\) and \(v(x)\). The product rule states:

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

Integrating both sides of this identity with respect to \(x\) yields:

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

Rewriting \(u' \, dx\) as \(du\) and \(v' \, dx\) as \(dv\) gives:

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

You can rearrange the equation. This isolates \(\int u \, dv\). As a result, you produce the integration by parts formula:

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

Every integration by parts problem is the product rule used in reverse. Similarly, u-substitution is the chain rule used in reverse.

Geometric interpretation: Imagine plotting two increasing functions. Let \(u\) and \(v\) be these functions. Plot them on a Cartesian coordinate system. Place \(v\) on the horizontal axis and \(u\) on the vertical axis. The rectangle formed from the origin \((0,0)\) to the point \((v, u)\) has an overall area of \(uv\). The curve divides this rectangle into two regions. One area is under the curve relative to \(v\). This is \(\int u \, dv\). The other area is to the left of the curve relative to \(u\). This is \(\int v \, du\). Adding these two areas together fills the rectangle \(uv\). This excludes any initial bounding box. As a result, this illustrates that the total area equals \(uv - \int v \, du\).

How do you choose u using the LIATE rule?

Choosing which function to set as \(u\) and which as \(dv\) is important. This choice determines if integration by parts simplifies the problem. Alternatively, it can make it harder. The LIATE rule is the standard guide for choosing \(u\). Select \(u\) as the function type that appears highest on this list. Then, let \(dv\) be everything else remaining in the integrand. Be sure to include \(dx\).

1. Logarithmic functions: \(\ln x\), \(\log_b x\)
2. Inverse trigonometric functions: \(\arctan x\), \(\arcsin x\), \(\arccos x\)
3. Algebraic functions: \(x\), \(x^2\), \(3x + 1\), polynomials, roots like \(\sqrt{x}\)
4. Trigonometric functions: \(\sin x\), \(\cos x\), \(\sec^2 x\)
5. Exponential functions: \(e^x\), \(e^{-2x}\), \(3^x\)

The strategic objective behind LIATE is clear. You should simplify \(u\) when you differentiate it to get \(du\). Also, \(dv\) should be straightforward to integrate into \(v\). It should not grow more complex.

Logarithmic and inverse trigonometric functions sit at the top of LIATE. They lack direct elementary antiderivatives. However, their derivatives turn into manageable rational algebraic expressions. Exponential and trigonometric functions sit at the bottom. They repeat or cycle predictably when integrated. Therefore, they are ideal choices for \(dv\).

When liate rule does not work

LIATE vs. ILATE: You may occasionally see this mnemonic written as ILATE. This places inverse trig functions before logarithms. However, both rules yield identical choices for nearly every problem. Integrands containing both a logarithm and an inverse trig function are extremely rare in standard calculus.

Exceptions to LIATE: LIATE is a rule of thumb. It is not a mathematical law. It assumes you can integrate \(dv\) on its own. For example, consider \(\int x^3 e^{x^2} dx\). Following LIATE strictly suggests you set \(u = x^3\). This is an algebraic term. Also set \(dv = e^{x^2} dx\). This is an exponential term. However, you cannot integrate \(e^{x^2}\) using elementary functions. The correct strategy splits the term. You can set \(u = x^2\) and \(dv = x e^{x^2} dx\). This is because you can easily integrate \(x e^{x^2}\) via substitution. You will get \(v = \frac{1}{2} e^{x^2}\). If LIATE leads to an impossible \(dv\), swap your choices. Similarly, swap them if it leads to a harder integral.

Step-by-step guide to integration by parts

Follow these five steps to solve any integration by parts problem systematically:

1. Assign \(u\) and \(dv\): Use the LIATE rule to choose \(u\). Then set \(dv\) equal to the rest of the integrand. Be sure to include \(dx\).
2. Compute \(du\) and \(v\): Differentiate \(u\) to find \(du = u'(x)dx\). Integrate \(dv\) to find \(v = \int dv\). (You do not need to add a constant \(+ C\) when finding \(v\)).
3. Apply the formula: Substitute \(u\), \(v\), \(du\), and \(dv\) into \(\int u \, dv = uv - \int v \, du\).
4. Evaluate the remaining integral: Integrate \(\int v \, du\). If this new integral still contains a product, perform a second round of parts. You can also use substitution, or apply basic integration formulas.
5. Add \(+ C\) and verify: Add the constant of integration \(+ C\) for indefinite integrals. Verify your answer by taking its derivative with the product rule to confirm it yields the original integrand.

Worked examples of integration by parts

Example 1: Algebraic and trigonometric functions — \(\int x \sin x \, dx\)

Following LIATE, the algebraic term \(x\) comes before the trigonometric term \(\sin x\). Set:

$$u = x \quad \implies \quad du = dx$$

$$dv = \sin x \, dx \quad \implies \quad v = -\cos x$$

Substitute these components into the formula:

$$\int x \sin x \, dx = -x \cos x - \int (-\cos x) dx$$. This simplifies to $$-x \cos x + \int \cos x \, dx$$. Therefore, you get $$\sin x - x \cos x + C$$.

If you mistakenly picked \(u = \sin x\) and \(dv = x \, dx\), you would get \(du = \cos x \, dx\). You would also get \(v = \frac{x^2}{2}\). As a result, the new integral would be \(\int \frac{x^2}{2} \cos x \, dx\). This integral has a higher power of \(x\). It is noticeably harder than the original. When the new integral becomes more complex, reverse your choices.

Example 2: Variant with cosine — \(\int x \cos x \, dx\)

Using \(u = x\) (\(du = dx\)) and \(dv = \cos x \, dx\) (\(v = \sin x\)):

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

Example 3: Single logarithmic function — \(\int \ln x \, dx\)

You can integrate single functions without an obvious second term. To do this, set \(dv = dx\). Here, the LIATE rule places logarithmic functions first.

$$u = \ln x \quad \implies \quad du = \frac{1}{x} dx$$

$$dv = dx \quad \implies \quad v = x$$

Applying the formula gives:

$$\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx$$. Next, you can simplify this to $$x \ln x - \int 1 \, dx$$. So, the result is $$x \ln x - x + C$$.

See the full walkthrough in integral of ln x.

Example 4: Single inverse trigonometric function — \(\int \arctan x \, dx\)

Set \(u = \arctan x\) and \(dv = dx\):

$$u = \arctan x \quad \implies \quad du = \frac{1}{1 + x^2} dx$$

$$dv = dx \quad \implies \quad v = x$$

Substitute into the formula:

$$\int \arctan x \, dx = x \arctan x - \int \frac{x}{1 + x^2} dx$$

The remaining integral requires a u-substitution. Let \(w = 1 + x^2\), so \(dw = 2x \, dx\), giving \(\int \frac{x}{1+x^2} dx = \frac{1}{2} \ln(1 + x^2)\). Combining terms yields:

$$x \arctan x - \frac{1}{2} \ln(1 + x^2) + C$$

Example 5: Inverse sine function — \(\int \arcsin x \, dx\)

Just like arctangent, choose \(u = \arcsin x\) and \(dv = dx\). Differentiating gives \(du = \frac{1}{\sqrt{1-x^2}} dx\) and integrating \(dv\) gives \(v = x\):

$$\int \arcsin x \, dx = x \arcsin x - \int \frac{x}{\sqrt{1 - x^2}} dx$$

Using substitution with \(w = 1 - x^2\) turns the second integral into \(-\sqrt{1-x^2}\). Subtracting this term completes the solution:

$$\int \arcsin x \, dx = x \arcsin x + \sqrt{1 - x^2} + C$$

Example 6: Product of algebraic and logarithmic functions — \(\int x \ln x \, dx\)

LIATE gives logarithmic terms priority over algebraic terms, so \(u = \ln x\) and \(dv = x \, dx\):

$$u = \ln x \quad \implies \quad du = \frac{1}{x} dx$$

$$dv = x \, dx \quad \implies \quad v = \frac{x^2}{2}$$

Applying the formula:

$$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx$$. This becomes $$\frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx$$. As a result, you find $$\frac{x^2}{2} \ln x - \frac{x^2}{4} + C$$.

The \(\frac{1}{x}\) factor comes from differentiating the logarithm. This factor cancels a power of \(x\). As a result, the problem becomes a simple polynomial integral.

Example 7: Repeated integration by parts — \(\int x^2 \cos x \, dx\)

When the polynomial factor has a power $n > 1$, you must perform integration by parts \(n\) times. First round: \(u = x^2\) and \(dv = \cos x \, dx\), which yields \(du = 2x \, dx\) and \(v = \sin x\):

$$\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx$$

The new integral \(\int 2x \sin x \, dx\) still contains a product of \(x\) and a trig function. So, run parts a second time on this sub-integral. Let \(u = 2x\) (\(du = 2 \, dx\)) and \(dv = \sin x \, dx\) (\(v = -\cos x\)).

$$\int 2x \sin x \, dx = -2x \cos x - \int (-2 \cos x) dx$$. You can then write this as $$-2x \cos x + 2 \sin x$$.

Substitute this back into the first equation, distributing the negative sign carefully:

$$\int x^2 \cos x \, dx = x^2 \sin x - (-2x \cos x + 2 \sin x) + C$$. This simplifies to $$x^2 \sin x + 2x \cos x - 2 \sin x + C$$.

Example 8: Cyclic or “looping” integrals — \(\int e^x \cos x \, dx\)

When integrating a product of an exponential function and a trigonometric function, neither factor simplifies to zero under differentiation. Set \(I = \int e^x \cos x \, dx\). First round: let \(u = \cos x\) and \(dv = e^x \, dx\), giving \(du = -\sin x \, dx\) and \(v = e^x\):

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

Run parts a second time on \(\int e^x \sin x \, dx\). Maintain consistent function assignments. Let \(u = \sin x\) (\(du = \cos x \, dx\)) and \(dv = e^x \, dx\) (\(v = e^x\)).

$$\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - I$$

Substitute this back into the main expression:

$$I = e^x \cos x + e^x \sin x - I$$

Notice that \(I\) appears on both sides. Add \(I\) to both sides to solve algebraically:

Therefore, $$2I = e^x \cos x + e^x \sin x$$. This implies that $$I = \frac{e^x(\sin x + \cos x)}{2} + C$$.

Maintain consistent choices across both rounds. If you swap roles in the second round, the terms cancel out. For example, choosing \(u = e^x\) instead of \(u = \sin x\) leaves \(I = I\). This undoes your progress.

How to evaluate definite and improper integrals

For definite integrals, evaluate the boundary terms \([uv]_a^b\) before computing the second integral:

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

Definite example 1: \(\int_0^{\pi} x \sin x \, dx\). Using \(u = x\) and \(dv = \sin x \, dx\):

$$\int_0^{\pi} x \sin x \, dx = \Big[-x \cos x\Big]_0^{\pi} - \int_0^{\pi} (-\cos x) dx$$. As a result, this equals $$\Big[-x \cos x + \sin x\Big]_0^{\pi}$$.

Evaluate at upper limit \(\pi\) and lower limit \(0\):

$$\Big(-\pi \cos(\pi) + \sin(\pi)\Big) - \Big(-0 \cos(0) + \sin(0)\Big) = (-\pi(-1) + 0) - (0) = \pi$$

Definite example 2: \(\int_1^e \ln x \, dx\). Using \(u = \ln x\) and \(dv = dx\):

$$\int_1^e \ln x \, dx = \Big[x \ln x - x\Big]_1^e$$. First, calculate $$(e \ln e - e)$$. Then subtract $$(1 \ln 1 - 1)$$. This gives $$(e(1) - e) - (0 - 1)$$, which simplifies to $$0 - (-1) = 1$$.

Improper integrals: Parts also handles improper integrals. For \(\int_0^{\infty} x e^{-x} \, dx\), set \(u = x\) and \(dv = e^{-x} dx\). This gives \(du = dx\) and \(v = -e^{-x}\).

$$\int_0^{\infty} x e^{-x} \, dx = \lim_{t \to \infty} \left( \Big[-x e^{-x}\Big]_0^t + \int_0^t e^{-x} dx \right)$$

The boundary term \([-x e^{-x}]_0^t = -t e^{-t} - 0\). As \(t \to \infty\), \(t e^{-t} = \frac{t}{e^t} \to 0\). This occurs by L’Hôpital’s rule. Exponential decay outpaces linear growth. Evaluating the remaining integral leaves \(\int_0^{\infty} e^{-x} \, dx = \Big[-e^{-x}\Big]_0^{\infty} = 0 - (-1) = 1\).

The tabular method for fast integration by parts

When you need to integrate a product of a high-degree polynomial and an exponential or trigonometric function, traditional integration by parts requires many repetitive steps. The tabular method (also known as the DI method or Stand-and-Deliver method) streamlines this process into a simple table.

To set up the tabular method:

1. Create two columns: a D (Differentiate) column and an I (Integrate) column.
2. Place the polynomial in the D column and repeatedly differentiate it until you reach 0.
3. Place the exponential or trigonometric function in the I column and integrate it the same number of times.
4. Assign alternating signs (\(+, -, +, -\)) starting with \(+\) next to the first row.
5. Multiply diagonally from each entry in column D. Go to the row below it in column I. Also, apply the assigned sign to each product.

Tabular example 1: Polynomial \(\times\) Exponential — \(\int x^2 e^x \, dx\)

List derivatives of \(x^2\) (\(x^2, 2x, 2, 0\)) and integrals of \(e^x\) (\(e^x, e^x, e^x, e^x\)). Multiplying diagonally with alternating signs gives:

$$\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2 e^x + C$$

More on this in integral of x·eˣ.

Tabular example 2: Polynomial \(\times\) Trigonometric — \(\int x^3 \sin x \, dx\)

The derivative column entries are \(x^3, 3x^2, 6x, 6, 0\). The integral column entries starting from \(\sin x\) are \(-\cos x, -\sin x, \cos x, \sin x\). Multiplying diagonally with alternating signs (\(+, -, +, -\)):

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

$$= -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C$$

The tabular method organizes repeated integration by parts concisely. It is most effective when one factor is a polynomial. This polynomial must differentiate to zero.

Common mistakes and how to fix them

• Differentiating \(dv\) instead of integrating it: A frequent error is computing \(v\) as the derivative of \(dv\). For instance, if \(dv = \sin x \, dx\), its antiderivative is \(v = -\cos x\), not \(\cos x\). Always double-check whether you are taking a derivative or an integral for each term.
• Minus sign errors in multi-step problems: In the formula \(uv - \int v \, du\), the minus sign applies to the entire second integral. When you perform a second round of integration by parts, enclose the entire new result in brackets. This ensures you distribute the outer negative sign correctly.
• Dropping the boundary evaluation in definite integrals: Evaluating \([uv]_a^b\) only at the upper limit leads to incorrect numerical values. Forgetting it entirely also causes errors. Compute \([uv]_a^b = u(b)v(b) - u(a)v(a)\) completely. Then move to the remaining integral.
• Switching choices in cyclic integrals: In looping problems like \(\int e^x \cos x \, dx\), if you choose the trigonometric function for \(u\) in round 1, you must choose the trigonometric function for \(u\) in round 2. Switching back to the exponential function undoes the integration and yields \(0 = 0\).
• Adding \(+ C\) too early or forgetting it entirely: Do not add a constant of integration when you find \(v\) from \(dv\). This is because constant additions cancel out during subtraction. Add a single \(+ C\) only after you evaluate the final indefinite integral.

When to use integration by parts vs. u-substitution

Knowing whether to use u-substitution or integration by parts saves time and prevents dead-end calculations. Use this comparison table to decide:

Feature	u-Substitution	Integration by Parts
Integrand structure	Composite function multiplied by inner derivative \(f(g(x))g'(x)\)	Product of two distinct function types \(f(x)g(x)\) or single log/inverse trig
Core mechanism	Reverse chain rule	Reverse product rule
Primary target	Simplifying composite terms like \((x^2+1)^5\) or \(e^{x^2}\)	Reducing algebraic powers or removing transcendental terms
Typical examples	\(\int 2x e^{x^2} dx\), \(\int \frac{\ln x}{x} dx\)	\(\int x e^x dx\), \(\int x \cos x dx\), \(\int \ln x dx\)

Applications across calculus and physics: Integration by parts is fundamental across advanced mathematics. It provides the standard antiderivatives for \(\ln x\), \(\arctan x\), and \(\arcsin x\). It also establishes reduction formulas for expressions like \(\int \sin^n x \, dx\). In probability theory, it computes expected values of continuous random variables. For example, it computes \(\int_0^{\infty} x e^{-x} dx\). It generates the integral remainder term in Taylor’s theorem. Furthermore, it is critical in engineering for deriving Fourier series coefficients and taking Laplace transforms of derivatives.

Try it yourself

You can check any antiderivative with the integral calculator, which shows the choice of \(u\) and \(dv\) in its working.

Practice problems

Test your understanding with these practice problems before checking the answers below.

1. \(\int x \cos(2x) \, dx\)
2. \(\int x^3 \ln x \, dx\)
3. \(\int_0^1 \arctan x \, dx\)
4. \(\int x e^{3x} \, dx\)
5. \(\int (\ln x)^2 \, dx\)
6. \(\int_0^{\pi/2} x \cos x \, dx\)

Answers and hints:

1. \(\frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} + C\) (Set \(u = x\) and \(dv = \cos(2x)dx\)).

2. \(\frac{x^4 \ln x}{4} - \frac{x^4}{16} + C\) (Set \(u = \ln x\) and \(dv = x^3 dx\)).

3. \(\frac{\pi}{4} - \frac{\ln 2}{2}\) (Set \(u = \arctan x\) and \(dv = dx\); evaluate from \(0\) to \(1\)).

4. \(\frac{x e^{3x}}{3} - \frac{e^{3x}}{9} + C\) (Set \(u = x\) and \(dv = e^{3x}dx\)).

5. \(x(\ln x)^2 - 2x \ln x + 2x + C\) (Set \(u = (\ln x)^2\) and \(dv = dx\). The remaining integral is \(2\int \ln x \, dx\)).

6. \(\frac{\pi}{2} - 1\) (Evaluates to \([x \sin x + \cos x]_0^{\pi/2} = (\frac{\pi}{2}(1) + 0) - (0 + 1) = \frac{\pi}{2} - 1\)).

Frequently asked questions

How do I know parts is the right method?

First, try u-substitution. If no part of the integrand is the derivative of another part, use integration by parts. Also use it if you have a product of different function types or a single inverse/log function.

Why is it called “by parts”?

You split the integrand into two distinct parts (\(u\) and \(dv\)), integrating one part while differentiating the other.

Does LIATE always work?

It works for the large majority of textbook problems. However, it does not work for all. For example, consider \(\int x^3 e^{x^2} dx\). The better split is \(u = x^2\) and \(dv = x e^{x^2} dx\). This is because you cannot integrate \(e^{x^2}\) alone in elementary terms.

Why isn’t a constant added when finding v?

You could add a constant \(K\) to get \(v + K\), but it cancels out completely in the formula. The extra \(K u\) in the \(uv\) term is subtracted by the additional \(\int K \, du\) in the second integral.

What if integration by parts keeps going forever?

Either you chose \(u\) and \(dv\) incorrectly, or you are dealing with a cyclic case. Incorrect choices make the new integral progressively harder. For example, consider \(\int e^x \cos x \, dx\). In this cyclic case, the original integral reappears. You must solve it algebraically.

Further reading

• Integration by Parts (Paul’s Online Math Notes) — the derivation plus a long set of worked examples, including the tabular shortcut.
• Integration Strategy (Paul’s Online Math Notes) — how to decide between parts, substitution and other techniques.