Integration by parts: formula, choices and worked examples

Integration by parts reverses the product rule for differentiation. It is mainly used when an integrand is a product and differentiating one factor makes it simpler.

The formula is

$$\boxed{\int u\,\frac{dv}{dx}\,dx=uv-\int v\,\frac{du}{dx}\,dx}.$$

It is often written more compactly as

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

The aim is not merely to apply the formula. It is to choose $u$ and $dv$ so that the new integral is easier than the original one.

Before you begin

You should be able to:

• differentiate products using the product rule
• differentiate and integrate powers, exponentials and trigonometric functions
• differentiate $\ln x$ and recognise that $\int 1\,dx=x$
• evaluate definite integrals
• distinguish integration by parts from integration by substitution

Where the formula comes from

For functions $u(x)$ and $v(x)$, the product rule says

$$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}.$$

Integrate every term with respect to $x$:

$$uv=\int u\frac{dv}{dx}\,dx+\int v\frac{du}{dx}\,dx.$$

Rearranging gives

$$\int u\frac{dv}{dx}\,dx =uv-\int v\frac{du}{dx}\,dx.$$

This derivation explains the minus sign. It also explains why one factor is differentiated while the other is integrated.

The practical method

Given an integral that can be regarded as a product:

1. Choose $u$, the factor to differentiate.
2. Let the rest of the integrand, including $dx$, be $dv$.
3. Find $du$ by differentiating $u$ and find $v$ by integrating $dv$.
4. Substitute into $\int u\,dv=uv-\int v\,du$.
5. Simplify and check whether the remaining integral is easier.

A small table keeps the roles clear:

$$\begin{array}{c|c} u=\cdots & dv=\cdots\,dx\\ du=\cdots\,dx & v=\cdots \end{array}$$

Only $v$ is an antiderivative. Do not include an arbitrary constant when finding $v$ inside the method. Any such constant cancels, and the final $+C$ supplies the full family of antiderivatives.

Choosing $u$ and $dv$

A good choice usually has both of these features:

• differentiating $u$ simplifies it
• integrating $dv$ is possible immediately

For a polynomial multiplied by $e^{ax}$, $\sin(ax)$ or $\cos(ax)$, choose the polynomial as $u$. Each differentiation lowers its degree.

For $\ln x$ or an inverse trigonometric function, choose that function as $u$ and treat the integrand as a product with $1$. These functions are easy to differentiate but do not have immediate standard antiderivatives at this stage.

The mnemonic LIATE can help choose $u$:

1. logarithmic functions
2. inverse trigonometric functions
3. algebraic functions, such as polynomials
4. trigonometric functions
5. exponential functions

Choose as $u$ the type appearing earliest in this list. LIATE is a guide, not a theorem. The decisive test is whether the new integral is simpler.

Parts or substitution?

Not every product requires integration by parts. First look for a composite function accompanied by its derivative.

$$\int 2x\cos(x^2)\,dx$$

is best handled by substitution because $2x$ is the derivative of $x^2$. In contrast,

$$\int x\cos x\,dx$$

has no inner function whose derivative is the other factor, and differentiating $x$ simplifies it. Integration by parts is natural.

A polynomial times an exponential

Worked example 1: $\int xe^x\,dx$

Choose $u=x$ because differentiating it gives $1$. Let $dv=e^x\,dx$ because $e^x$ is easy to integrate.

$$\begin{array}{c|c} u=x & dv=e^x\,dx\\ du=dx & v=e^x \end{array}$$

Apply the formula:

$$\begin{aligned} \int xe^x\,dx &=xe^x-\int e^x\,dx\\ &=xe^x-e^x+C\\ &=\boxed{e^x(x-1)+C}. \end{aligned}$$

Differentiate to check:

$$\frac{d}{dx}\left[e^x(x-1)\right] =e^x(x-1)+e^x =xe^x.$$

Worked example 2: a non-unit coefficient

Find

$$\int x e^{3x}\,dx.$$

Use $u=x$ and $dv=e^{3x}\,dx$. Take care when integrating the exponential:

$$du=dx, \qquad v=\frac13e^{3x}.$$

Therefore

$$\begin{aligned} \int xe^{3x}\,dx &=\frac{x}{3}e^{3x}-\int\frac13e^{3x}\,dx\\ &=\frac{x}{3}e^{3x}-\frac19e^{3x}+C\\ &=\boxed{\frac{e^{3x}}9(3x-1)+C}. \end{aligned}$$

The factor $1/3$ appears once when finding $v$ and again when integrating $e^{3x}$ in the remaining integral.

A polynomial times a trigonometric function

Worked example 3: $\int x\cos x\,dx$

Let

$$u=x, \qquad dv=\cos x\,dx.$$

Then

$$du=dx, \qquad v=\sin x.$$

Hence

$$\begin{aligned} \int x\cos x\,dx &=x\sin x-\int\sin x\,dx\\ &=\boxed{x\sin x+\cos x+C}. \end{aligned}$$

The final sign is positive because $\int\sin x\,dx=-\cos x$, and the formula already contains a subtraction.

Worked example 4: signs and coefficients

Find

$$\int x\sin(2x)\,dx.$$

Choose $u=x$ and $dv=\sin(2x)\,dx$. Then

$$du=dx, \qquad v=-\frac12\cos(2x).$$

Thus

$$\begin{aligned} \int x\sin(2x)\,dx &=-\frac{x}{2}\cos(2x) -\int\left(-\frac12\cos(2x)\right)dx\\ &=-\frac{x}{2}\cos(2x)+\frac12\int\cos(2x)\,dx\\ &=\boxed{-\frac{x}{2}\cos(2x)+\frac14\sin(2x)+C}. \end{aligned}$$

Repeated integration by parts

If $u$ is a polynomial of degree $2$ or more, one application usually leaves another product requiring integration by parts. Keep differentiating the polynomial until it becomes zero.

Worked example 5: two applications

Find

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

First use $u=x^2$ and $dv=e^x\,dx$:

$$\begin{aligned} \int x^2e^x\,dx &=x^2e^x-\int 2xe^x\,dx\\ &=x^2e^x-2\int xe^x\,dx. \end{aligned}$$

The remaining integral is Example 1, or can be handled by parts again:

$$\int xe^x\,dx=xe^x-e^x.$$

Therefore

$$\begin{aligned} \int x^2e^x\,dx &=x^2e^x-2(xe^x-e^x)+C\\ &=\boxed{e^x(x^2-2x+2)+C}. \end{aligned}$$

There is only one arbitrary constant, written after the whole calculation.

The tabular shortcut

For repeated products of a polynomial with a repeatedly integrable function, the same work can be organised as follows:

$$\begin{array}{c|c|c} \text{sign} & \text{differentiate }x^2 & \text{integrate }e^x\\ \hline + & x^2 & e^x\\ - & 2x & e^x\\ + & 2 & e^x\\ - & 0 & e^x \end{array}$$

Multiply diagonally and add:

$$x^2e^x-2xe^x+2e^x+C.$$

The tabular method is compressed repeated integration by parts. It should not replace understanding of the formula.

Integrating logarithms

An expression such as $\ln x$ is secretly a product:

$$\ln x=1\cdot\ln x.$$

Choose $u=\ln x$, since it becomes the simpler function $1/x$ when differentiated, and choose $dv=dx$.

Worked example 6: $\int\ln x\,dx$

For $x>0$,

$$u=\ln x, \qquad dv=dx, \qquad du=\frac1x\,dx, \qquad v=x.$$

Then

$$\begin{aligned} \int\ln x\,dx &=x\ln x-\int x\left(\frac1x\right)dx\\ &=x\ln x-\int1\,dx\\ &=\boxed{x\ln x-x+C}. \end{aligned}$$

More generally, on any interval not crossing $x=0$,

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

Worked example 7: an algebraic factor with a logarithm

Find

$$\int x\ln x\,dx, \qquad x>0.$$

LIATE suggests choosing $u=\ln x$. Put the remaining factor into $dv$:

$$u=\ln x, \quad dv=x\,dx, \quad du=\frac1x\,dx, \quad v=\frac{x^2}{2}.$$

Therefore

$$\begin{aligned} \int x\ln x\,dx &=\frac{x^2}{2}\ln x -\int\frac{x^2}{2}\frac1x\,dx\\ &=\frac{x^2}{2}\ln x-\frac12\int x\,dx\\ &=\boxed{\frac{x^2}{2}\ln x-\frac{x^2}{4}+C}. \end{aligned}$$

Choosing $u=x$ would force you to integrate $\ln x$ before you had solved that problem, so it does not simplify the work.

Definite integrals

For limits $a$ and $b$, integration by parts becomes

$$\boxed{\int_a^b u\,dv=[uv]_a^b-\int_a^b v\,du}.$$

Keep the limits on the remaining integral and evaluate the boundary term at both endpoints. An alternative is to find an indefinite antiderivative first and then apply the limits. Do not mix the two approaches halfway through.

Worked example 8: definite integration by parts

Evaluate

$$\int_0^1 xe^{2x}\,dx.$$

Let $u=x$ and $dv=e^{2x}\,dx$. Then $du=dx$ and $v=\frac12e^{2x}$.

$$\begin{aligned} \int_0^1xe^{2x}\,dx &=\left[\frac{x}{2}e^{2x}\right]_0^1 -\frac12\int_0^1e^{2x}\,dx\\ &=\frac{e^2}{2}-\frac12\left[\frac12e^{2x}\right]_0^1\\ &=\frac{e^2}{2}-\frac14(e^2-1)\\ &=\boxed{\frac{e^2+1}{4}}. \end{aligned}$$

The value is positive, as expected because $xe^{2x}\geq0$ on $[0,1]$.

Worked example 9: logarithm with an endpoint

Evaluate

$$\int_1^e\ln x\,dx.$$

Using $\int\ln x\,dx=x\ln x-x$,

$$\begin{aligned} \int_1^e\ln x\,dx &=[x\ln x-x]_1^e\\ &=(e-e)-(0-1)\\ &=\boxed{1}. \end{aligned}$$

When the original integral returns

For some products, applying integration by parts twice produces the original integral. This is useful: collect that integral algebraically and solve for it.

Worked example 10: $\int e^x\cos x\,dx$

Let

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

Choose $u=\cos x$ and $dv=e^x\,dx$:

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

Apply integration by parts to the new integral, choosing $u=\sin x$ and $dv=e^x\,dx$:

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

Substitute this back:

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

Now solve for $I$:

$$2I=e^x(\sin x+\cos x),$$

so

$$\boxed{\int e^x\cos x\,dx =\frac12e^x(\sin x+\cos x)+C}.$$

Do not continue applying integration by parts in a loop. As soon as the original integral reappears, treat it as an algebraic unknown.

Common misconceptions

Choosing $dv$ that cannot be integrated

For $\int x\ln x\,dx$, choosing $u=x$ and $dv=\ln x\,dx$ is unhelpful because finding $v$ is already the main problem. Choose $u=\ln x$ instead.

Forgetting the minus sign

The formula is

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

not $uv+\int v\,du$. Derive it quickly from the product rule if uncertain.

Differentiating both factors

Integration by parts does not replace $\int fg\,dx$ with $\int f'g'\,dx$. One chosen part is differentiated and the other is integrated.

Dropping chain rule factors

If $dv=e^{4x}\,dx$, then $v=\frac14e^{4x}$. If $dv=\sin(3x)\,dx$, then $v=-\frac13\cos(3x)$. Differentiate $v$ mentally to check it returns $dv/dx$.

Assuming every product needs parts

The integral $\int x(x^2+1)^5\,dx$ is better handled by substitution because the derivative of $x^2+1$ is proportional to $x$.

Mishandling definite limits

In $[uv]_a^b$, both $u$ and $v$ are evaluated at both limits. The remaining integral also retains the limits $a$ and $b$.

Self-check

Try each question before opening its answer.

1. One application

Find

$$\int x\cos(3x)\,dx.$$

Answer

Take $u=x$ and $dv=\cos(3x)\,dx$. Then $du=dx$ and $v=\frac13\sin(3x)$.

$$\begin{aligned} \int x\cos(3x)\,dx &=\frac{x}{3}\sin(3x)-\frac13\int\sin(3x)\,dx\\ &=\boxed{\frac{x}{3}\sin(3x)+\frac19\cos(3x)+C}. \end{aligned}$$

2. Repeated use

Find

$$\int x^2\sin x\,dx.$$

Answer

First choose $u=x^2$ and $dv=\sin x\,dx$:

$$\begin{aligned} \int x^2\sin x\,dx &=-x^2\cos x+2\int x\cos x\,dx\\ &=-x^2\cos x+2(x\sin x+\cos x)+C\\ &=\boxed{-x^2\cos x+2x\sin x+2\cos x+C}. \end{aligned}$$

3. Logarithmic function

Find, for $x>0$,

$$\int x^3\ln x\,dx.$$

Answer

Choose $u=\ln x$ and $dv=x^3\,dx$. Then $du=\frac1x\,dx$ and $v=\frac{x^4}{4}$.

$$\begin{aligned} \int x^3\ln x\,dx &=\frac{x^4}{4}\ln x-\frac14\int x^3\,dx\\ &=\boxed{\frac{x^4}{4}\ln x-\frac{x^4}{16}+C}. \end{aligned}$$

4. Definite integral

Evaluate

$$\int_0^{\pi/2}x\sin x\,dx.$$

Answer

Take $u=x$ and $dv=\sin x\,dx$, so $du=dx$ and $v=-\cos x$.

$$\begin{aligned} \int_0^{\pi/2}x\sin x\,dx &=[-x\cos x]_0^{\pi/2}+\int_0^{\pi/2}\cos x\,dx\\ &=0+[\sin x]_0^{\pi/2}\\ &=\boxed{1}. \end{aligned}$$

5. Choose the method

State the best main method for each integral.

$$\text{(a) }\int xe^{x^2}\,dx, \qquad \text{(b) }\int x^2e^x\,dx, \qquad \text{(c) }\int\ln x\,dx.$$

Answer

• (a) Substitution, because $x$ is proportional to the derivative of the exponent $x^2$.
• (b) Repeated integration by parts, choosing $u=x^2$.
• (c) Integration by parts, writing the integrand as $1\cdot\ln x$ and choosing $u=\ln x$.

Summary

The essential facts are:

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

• choose $u$ so that differentiating it simplifies the expression
• choose $dv$ so that it can be integrated immediately
• expect repeated applications when $u$ is a polynomial of degree greater than $1$
• write logarithms as products with $1$
• preserve limits throughout a definite integral
• if the original integral returns, collect it algebraically
• verify an indefinite answer by differentiation

Next, compare this technique with integration by substitution and learn how partial fractions prepare rational functions for integration. For mixed questions, practise choosing an appropriate method.