1.11.5 Integration by substitution and by parts (A-level only)

What you'll learn

How integration by substitution reverses the chain rule.
How to choose a suitable substitution and transform an integral completely.
How integration by parts reverses the product rule.
How to apply integration by parts more than once when necessary.

Before you begin: reversing differentiation

Integration is the inverse process of differentiation. This means that differentiation rules can suggest methods for integrating more complicated expressions.

Two differentiation rules are especially important here:

The chain rule differentiates a composite function, such as $(3x+1)5\left(3x+1\right)^5$ .
The product rule differentiates a product of two functions, such as $x^2e^x$ .

Integration by substitution reverses the chain rule, while integration by parts reverses the product rule.

Integration by substitution

The connection with the chain rule

A composite function is one function applied to another. For example, in $sin⁡(x2)\sin\left(x^2\right)$ , the inner function is $x^2$ and the outer function is sine.

The chain rule gives

ddxF(g(x))=F′(g(x))g′(x).\frac{d}{dx}F\left(g(x)\right)=F'\left(g(x)\right)g'(x).

Reversing this tells you that

dx=F(g(x))+C.\int F'\left(g(x)\right)g'(x)\,dx=F\left(g(x)\right)+C.

Therefore, an integral containing a function and the derivative of its inner expression is a strong candidate for substitution.

Definition

Integration by substitution

Integration by substitution introduces a new variable, usually $u$ , to replace part of the integrand. The aim is to transform the integral into a simpler one that can be integrated directly.

Carrying out a substitution

If you choose

u = g (x),

then differentiate with respect to $x$ :

dudx=g′(x).\frac{du}{dx}=g'(x).

This can be rearranged as

dx.du=g'(x)\,dx.

You must rewrite the whole integral, including the differential $d x$ , in terms of the new variable.

Key Idea

What makes a useful substitution?

Choose an inner expression whose derivative also appears in the integral, possibly differing only by a constant factor. A successful substitution should leave an integral involving only the new variable.

Example

Integrating a power of a linear expression

Find

dx.\int x\left(x^2+4\right)^5\,dx.

Choose the inner expression $u=x^2+4$ . Differentiating gives $dxdu=2x\,dx$ , so $dx=12dux\,dx=\frac{1}{2}du$ .
Replace both parts of the integral:
$du.\int x\left(x^2+4\right)^5\,dx =\frac{1}{2}\int u^5\,du.$
Integrate with respect to $u$ :
$du=u612+C.\frac{1}{2}\int u^5\,du =\frac{u^6}{12}+C.$
Substitute back using $u=x^2+4$ :
$(x2+4)612+C.\boxed{\frac{\left(x^2+4\right)^6}{12}+C}.$

Common Mistake

Forgetting the derivative factor

You cannot simply replace $x^2+4$ by $u$ and ignore $dxx\,dx$ . The differential must also be transformed using $dxdu=2x\,dx$ .

Finding a suitable substitution

The question may not tell you what substitution to use. Look for an expression inside a power, root, exponential, denominator or trigonometric function.

Useful patterns include:

$f′(x)(f(x))nf'(x)\left(f(x)\right)^n$
$f′(x)f(x)\dfrac{f'(x)}{f(x)}$
$f'(x)e^{f(x)}$
$f′(x)cos⁡(f(x))f'(x)\cos\left(f(x)\right)$
$f′(x)sin⁡(f(x))f'(x)\sin\left(f(x)\right)$

A constant multiple of the required derivative is not a problem: you can adjust for it outside the integral.

Example

Choosing a substitution involving a denominator

Find

dx.\int \frac{x}{3x^2+5}\,dx.

The denominator is a natural choice, so let $u=3x^2+5$ . Then $dxdu=6x\,dx$ , giving $dx=16dux\,dx=\frac{1}{6}du$ .
Transform the integral:
$du.\int \frac{x}{3x^2+5}\,dx =\frac{1}{6}\int \frac{1}{u}\,du.$
Use $du=ln⁡∣u∣+C\int \frac{1}{u}\,du=\ln|u|+C$ :
$16ln⁡∣u∣+C.\frac{1}{6}\ln|u|+C.$
Substitute back:
$16ln⁡∣3x2+5∣+C.\boxed{\frac{1}{6}\ln\left|3x^2+5\right|+C}.$
Since $3x^2+5$ is always positive, this could also be written as $16ln⁡(3x2+5)+C\frac{1}{6}\ln\left(3x^2+5\right)+C$ .

Substitution in definite integrals

For a definite integral, you can transform the limits into values of $u$ . Once the limits have been changed, complete the calculation entirely in terms of $u$ and do not substitute back to $x$ .

Example

Changing the limits of a definite integral

Evaluate

dx.\int_0^1 2x\sqrt{x^2+3}\,dx.

Let $u=x^2+3$ , so $dxdu=2x\,dx$ . Transform the limits: when $x = 0$ , $u = 3$ ; when $x = 1$ , $u = 4$ .
Rewrite the integral using the new limits:
$du.\int_0^1 2x\sqrt{x^2+3}\,dx =\int_3^4 u^{1/2}\,du.$
Integrate and evaluate:
$du=[23u3/2]34=23(8−33)=163−23.\begin{aligned} \int_3^4 u^{1/2}\,du &=\left[\frac{2}{3}u^{3/2}\right]_3^4\\ &=\frac{2}{3}\left(8-3\sqrt{3}\right)\\ &=\boxed{\frac{16}{3}-2\sqrt{3}}. \end{aligned}$

Tip

Two valid approaches to definite integrals

Either change the limits to values of $u$ , or find an antiderivative and substitute back to $x$ before using the original limits. Do not mix the two approaches.

Integration by parts

The connection with the product rule

The product rule states

ddx(uv)=udvdx+vdudx.\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}.

Integrating both sides and rearranging produces the integration-by-parts formula.

Definition

Integration by parts

The formula for integration by parts is

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

It is often written more compactly as

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

The method is useful when the integrand is a product and differentiating one factor makes it simpler.

Choosing $u$ and $d v$

You split the integrand into two parts:

Choose $u$ to be the factor that becomes simpler when differentiated.
Choose $d v$ to be the factor that can be integrated easily.

A useful guide for choosing $u$ is LIATE:

Logarithmic functions, such as $ln⁡x\ln x$
Inverse trigonometric functions
Algebraic functions, such as $x$ or $x^2$
Trigonometric functions
Exponential functions

This is guidance, not an inflexible rule.

Example

Integrating an algebraic-exponential product

Find

dx.\int xe^x\,dx.

Choose $u = x$ because differentiating it makes it simpler, and choose $dv=e^x\,dx$ because it is easy to integrate. Then $d u = d x$ and $v=e^x$ .
Apply $du\int u\,dv=uv-\int v\,du$ :
$dx.\int xe^x\,dx=xe^x-\int e^x\,dx.$
Complete the remaining integral:
$xex−ex+C.\boxed{xe^x-e^x+C}.$
Differentiate the answer to check:
$ddx(xex−ex)=ex+xex−ex=xex.\frac{d}{dx}\left(xe^x-e^x\right) =e^x+xe^x-e^x =xe^x.$

Common Mistake

Losing the minus sign

The formula contains a subtraction: $duuv-\int v\,du$ . Write the formula before substituting your choices to reduce sign errors.

When there is no obvious product

Some integrals can be treated as products by inserting a factor of 1. This is particularly useful for logarithmic and inverse trigonometric functions.

Example

Integrating a logarithm

Find

dx.\int \ln x\,dx.

Regard the integrand as $1⋅ln⁡x1\cdot\ln x$ . Choose $u=ln⁡xu=\ln x$ and $d v = d x$ , giving $du=1xdxdu=\frac{1}{x}dx$ and $v = x$ .
Apply integration by parts:
$dx.\int \ln x\,dx =x\ln x-\int x\left(\frac{1}{x}\right)\,dx.$
Simplify and integrate:
$xln⁡x−x+C.\boxed{x\ln x-x+C}.$

Applying integration by parts more than once

If the remaining integral is still a product of the same general type, apply integration by parts again. This is common when a polynomial is multiplied by an exponential or trigonometric function.

Example

Applying integration by parts twice

Find

dx.\int x^2\cos x\,dx.

Choose $u=x^2$ and $dxdv=\cos x\,dx$ . Then $dxdu=2x\,dx$ and $v=sin⁡xv=\sin x$ , so
$dx.\int x^2\cos x\,dx =x^2\sin x-\int 2x\sin x\,dx.$
For the remaining integral, choose $u = 2 x$ and $dxdv=\sin x\,dx$ . Then $dxdu=2\,dx$ and $v=−cos⁡xv=-\cos x$ :
$dx.\int 2x\sin x\,dx =-2x\cos x+\int 2\cos x\,dx.$
Evaluate the final integral:
$dx=−2xcos⁡x+2sin⁡x.\int 2x\sin x\,dx=-2x\cos x+2\sin x.$
Substitute this into the result from the first application:
$dx=x2sin⁡x−(−2xcos⁡x+2sin⁡x)=x2sin⁡x+2xcos⁡x−2sin⁡x+C.\begin{aligned} \int x^2\cos x\,dx &=x^2\sin x-\left(-2x\cos x+2\sin x\right)\\ &=\boxed{x^2\sin x+2x\cos x-2\sin x+C}. \end{aligned}$

Key Idea

Know which method you are reversing

Use substitution when you can see a composite function together with its inner derivative. Use integration by parts when a product contains a factor that becomes simpler when differentiated.

Common Mistake

Not every substitution succeeds

At this level, substitution questions are limited to cases where one suitable substitution produces an integrable function. If both $x$ and $u$ remain after your substitution, reconsider your choice or use an algebraic relation to rewrite the remaining $x$ -terms.

Exam technique

In the exam

Decide whether the integral resembles a reversed chain rule or a reversed product rule.
For substitution, write $u$ , $d u$ and any changed limits explicitly; check that no $x$ remains.
For integration by parts, record $u$ , $d u$ , $d v$ and $v$ before using the formula.
Use brackets carefully when subtracting an integral produced by a second application of integration by parts.
Differentiate your final antiderivative when time allows, and include $+ C$ for every indefinite integral.

Self review

Check yourself

What substitution would you try for $dx\int x\cos\left(x^2\right)\,dx$ , and why?
How would you choose $u$ and $d v$ when integrating $dx\int x^3e^x\,dx$ ?
When changing the limits of a definite integral to values of $u$ , why should you not substitute back to $x$ ?

What you'll learn

How integration by substitution reverses the chain rule.
How to choose a suitable substitution and transform an integral completely.
How integration by parts reverses the product rule.
How to apply integration by parts more than once when necessary.

Before you begin: reversing differentiation

Integration is the inverse process of differentiation. This means that differentiation rules can suggest methods for integrating more complicated expressions.

Two differentiation rules are especially important here:

The chain rule differentiates a composite function, such as $(3x+1)5\left(3x+1\right)^5$ .
The product rule differentiates a product of two functions, such as $x^2e^x$ .

Integration by substitution reverses the chain rule, while integration by parts reverses the product rule.

Integration by substitution

The connection with the chain rule

A composite function is one function applied to another. For example, in $sin⁡(x2)\sin\left(x^2\right)$ , the inner function is $x^2$ and the outer function is sine.

The chain rule gives

ddxF(g(x))=F′(g(x))g′(x).\frac{d}{dx}F\left(g(x)\right)=F'\left(g(x)\right)g'(x).

Reversing this tells you that

dx=F(g(x))+C.\int F'\left(g(x)\right)g'(x)\,dx=F\left(g(x)\right)+C.

Therefore, an integral containing a function and the derivative of its inner expression is a strong candidate for substitution.

Definition

Integration by substitution

Integration by substitution introduces a new variable, usually $u$ , to replace part of the integrand. The aim is to transform the integral into a simpler one that can be integrated directly.

Carrying out a substitution

If you choose

u = g (x),

then differentiate with respect to $x$ :

dudx=g′(x).\frac{du}{dx}=g'(x).

This can be rearranged as

dx.du=g'(x)\,dx.

You must rewrite the whole integral, including the differential $d x$ , in terms of the new variable.

Key Idea

What makes a useful substitution?

Example

Integrating a power of a linear expression

Find

dx.\int x\left(x^2+4\right)^5\,dx.

Choose the inner expression $u=x^2+4$ . Differentiating gives $dxdu=2x\,dx$ , so $dx=12dux\,dx=\frac{1}{2}du$ .
Replace both parts of the integral:
$du.\int x\left(x^2+4\right)^5\,dx =\frac{1}{2}\int u^5\,du.$
Integrate with respect to $u$ :
$du=u612+C.\frac{1}{2}\int u^5\,du =\frac{u^6}{12}+C.$
Substitute back using $u=x^2+4$ :
$(x2+4)612+C.\boxed{\frac{\left(x^2+4\right)^6}{12}+C}.$

Common Mistake

Forgetting the derivative factor

You cannot simply replace $x^2+4$ by $u$ and ignore $dxx\,dx$ . The differential must also be transformed using $dxdu=2x\,dx$ .

Finding a suitable substitution

The question may not tell you what substitution to use. Look for an expression inside a power, root, exponential, denominator or trigonometric function.

Useful patterns include:

$f′(x)(f(x))nf'(x)\left(f(x)\right)^n$
$f′(x)f(x)\dfrac{f'(x)}{f(x)}$
$f'(x)e^{f(x)}$
$f′(x)cos⁡(f(x))f'(x)\cos\left(f(x)\right)$
$f′(x)sin⁡(f(x))f'(x)\sin\left(f(x)\right)$

A constant multiple of the required derivative is not a problem: you can adjust for it outside the integral.

Example

Choosing a substitution involving a denominator

Find

dx.\int \frac{x}{3x^2+5}\,dx.

The denominator is a natural choice, so let $u=3x^2+5$ . Then $dxdu=6x\,dx$ , giving $dx=16dux\,dx=\frac{1}{6}du$ .
Transform the integral:
$du.\int \frac{x}{3x^2+5}\,dx =\frac{1}{6}\int \frac{1}{u}\,du.$
Use $du=ln⁡∣u∣+C\int \frac{1}{u}\,du=\ln|u|+C$ :
$16ln⁡∣u∣+C.\frac{1}{6}\ln|u|+C.$
Substitute back:
$16ln⁡∣3x2+5∣+C.\boxed{\frac{1}{6}\ln\left|3x^2+5\right|+C}.$
Since $3x^2+5$ is always positive, this could also be written as $16ln⁡(3x2+5)+C\frac{1}{6}\ln\left(3x^2+5\right)+C$ .

Substitution in definite integrals

For a definite integral, you can transform the limits into values of $u$ . Once the limits have been changed, complete the calculation entirely in terms of $u$ and do not substitute back to $x$ .

Example

Changing the limits of a definite integral

Evaluate

dx.\int_0^1 2x\sqrt{x^2+3}\,dx.

Let $u=x^2+3$ , so $dxdu=2x\,dx$ . Transform the limits: when $x = 0$ , $u = 3$ ; when $x = 1$ , $u = 4$ .
Rewrite the integral using the new limits:
$du.\int_0^1 2x\sqrt{x^2+3}\,dx =\int_3^4 u^{1/2}\,du.$
Integrate and evaluate:
$du=[23u3/2]34=23(8−33)=163−23.\begin{aligned} \int_3^4 u^{1/2}\,du &=\left[\frac{2}{3}u^{3/2}\right]_3^4\\ &=\frac{2}{3}\left(8-3\sqrt{3}\right)\\ &=\boxed{\frac{16}{3}-2\sqrt{3}}. \end{aligned}$

Tip

Two valid approaches to definite integrals

Either change the limits to values of $u$ , or find an antiderivative and substitute back to $x$ before using the original limits. Do not mix the two approaches.

Integration by parts

The connection with the product rule

The product rule states

ddx(uv)=udvdx+vdudx.\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}.

Integrating both sides and rearranging produces the integration-by-parts formula.

Definition

Integration by parts

The formula for integration by parts is

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

It is often written more compactly as

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

The method is useful when the integrand is a product and differentiating one factor makes it simpler.

Choosing $u$ and $d v$

You split the integrand into two parts:

Choose $u$ to be the factor that becomes simpler when differentiated.
Choose $d v$ to be the factor that can be integrated easily.

A useful guide for choosing $u$ is LIATE:

Logarithmic functions, such as $ln⁡x\ln x$
Inverse trigonometric functions
Algebraic functions, such as $x$ or $x^2$
Trigonometric functions
Exponential functions

This is guidance, not an inflexible rule.

Example

Integrating an algebraic-exponential product

Find

dx.\int xe^x\,dx.

Choose $u = x$ because differentiating it makes it simpler, and choose $dv=e^x\,dx$ because it is easy to integrate. Then $d u = d x$ and $v=e^x$ .
Apply $du\int u\,dv=uv-\int v\,du$ :
$dx.\int xe^x\,dx=xe^x-\int e^x\,dx.$
Complete the remaining integral:
$xex−ex+C.\boxed{xe^x-e^x+C}.$
Differentiate the answer to check:
$ddx(xex−ex)=ex+xex−ex=xex.\frac{d}{dx}\left(xe^x-e^x\right) =e^x+xe^x-e^x =xe^x.$

Common Mistake

Losing the minus sign

The formula contains a subtraction: $duuv-\int v\,du$ . Write the formula before substituting your choices to reduce sign errors.

When there is no obvious product

Some integrals can be treated as products by inserting a factor of 1. This is particularly useful for logarithmic and inverse trigonometric functions.

Example

Integrating a logarithm

Find

dx.\int \ln x\,dx.

Regard the integrand as $1⋅ln⁡x1\cdot\ln x$ . Choose $u=ln⁡xu=\ln x$ and $d v = d x$ , giving $du=1xdxdu=\frac{1}{x}dx$ and $v = x$ .
Apply integration by parts:
$dx.\int \ln x\,dx =x\ln x-\int x\left(\frac{1}{x}\right)\,dx.$
Simplify and integrate:
$xln⁡x−x+C.\boxed{x\ln x-x+C}.$

Applying integration by parts more than once

If the remaining integral is still a product of the same general type, apply integration by parts again. This is common when a polynomial is multiplied by an exponential or trigonometric function.

Example

Applying integration by parts twice

Find

dx.\int x^2\cos x\,dx.

Choose $u=x^2$ and $dxdv=\cos x\,dx$ . Then $dxdu=2x\,dx$ and $v=sin⁡xv=\sin x$ , so
$dx.\int x^2\cos x\,dx =x^2\sin x-\int 2x\sin x\,dx.$
For the remaining integral, choose $u = 2 x$ and $dxdv=\sin x\,dx$ . Then $dxdu=2\,dx$ and $v=−cos⁡xv=-\cos x$ :
$dx.\int 2x\sin x\,dx =-2x\cos x+\int 2\cos x\,dx.$
Evaluate the final integral:
$dx=−2xcos⁡x+2sin⁡x.\int 2x\sin x\,dx=-2x\cos x+2\sin x.$
Substitute this into the result from the first application:
$dx=x2sin⁡x−(−2xcos⁡x+2sin⁡x)=x2sin⁡x+2xcos⁡x−2sin⁡x+C.\begin{aligned} \int x^2\cos x\,dx &=x^2\sin x-\left(-2x\cos x+2\sin x\right)\\ &=\boxed{x^2\sin x+2x\cos x-2\sin x+C}. \end{aligned}$

Key Idea

Know which method you are reversing

Use substitution when you can see a composite function together with its inner derivative. Use integration by parts when a product contains a factor that becomes simpler when differentiated.

Common Mistake

Not every substitution succeeds

Exam technique

In the exam

Decide whether the integral resembles a reversed chain rule or a reversed product rule.
For substitution, write $u$ , $d u$ and any changed limits explicitly; check that no $x$ remains.
For integration by parts, record $u$ , $d u$ , $d v$ and $v$ before using the formula.
Use brackets carefully when subtracting an integral produced by a second application of integration by parts.
Differentiate your final antiderivative when time allows, and include $+ C$ for every indefinite integral.

Self review

Check yourself

What substitution would you try for $dx\int x\cos\left(x^2\right)\,dx$ , and why?
How would you choose $u$ and $d v$ when integrating $dx\int x^3e^x\,dx$ ?
When changing the limits of a definite integral to values of $u$ , why should you not substitute back to $x$ ?

What you'll learn

Before you begin: reversing differentiation

Integration by substitution

The connection with the chain rule

Carrying out a substitution

Finding a suitable substitution

Substitution in definite integrals

Integration by parts

The connection with the product rule

Choosing $u$ and $d v$

When there is no obvious product

Applying integration by parts more than once

Review 1.11.5 Integration by substitution and by parts (A-level only) by teaching Genie

1.11.5 Integration by substitution and by parts (A-level only)

What you'll learn

Before you begin: reversing differentiation

Integration by substitution

The connection with the chain rule

Carrying out a substitution

Finding a suitable substitution

Substitution in definite integrals

Integration by parts

The connection with the product rule

Choosing $u$ and $d v$

When there is no obvious product

Applying integration by parts more than once

Review 1.11.5 Integration by substitution and by parts (A-level only) by teaching Genie

1.11.5 Integration by substitution and by parts (A-level only)

What you'll learn

Before you begin: reversing differentiation

Integration by substitution

The connection with the chain rule

Carrying out a substitution

Finding a suitable substitution

Substitution in definite integrals

Integration by parts

The connection with the product rule

Choosing uuu and dvdvdv

When there is no obvious product

Applying integration by parts more than once

Review 1.11.5 Integration by substitution and by parts (A-level only) by teaching Genie

1.11.5 Integration by substitution and by parts (A-level only) Revision Guide

1.11.5 Integration by substitution and by parts (A-level only)

What you'll learn

Before you begin: reversing differentiation

Integration by substitution

The connection with the chain rule

Carrying out a substitution

Finding a suitable substitution

Substitution in definite integrals

Integration by parts

The connection with the product rule

Choosing uuu and dvdvdv

When there is no obvious product

Applying integration by parts more than once

Review 1.11.5 Integration by substitution and by parts (A-level only) by teaching Genie

1.11.5 Integration by substitution and by parts (A-level only) Revision Guide

Choosing $u$ and $d v$

Choosing $u$ and $d v$