H5PureStretch

Integration by Substitution

Something like $\int 2 x (x^{2} + 1)^{5} d x$ has no obvious antiderivative from the power rule alone — it is a composite function in disguise. Substitution reverses the chain rule, replacing a complicated expression with a single new variable to make the integral solvable.

36 min Video by TLMaths Integration

The big picture

Just as the chain rule differentiates a function nested inside another, integration by substitution reverses that exact process. By spotting a function AND its own derivative sitting together inside the integral, you can rename the inner function $u$ and rewrite the entire integral purely in terms of $u$ — usually collapsing it down to a simple power-rule integral. Recognising when a substitution will work (and choosing the right one) is the key skill.

A-Level Maths H5-12: Further Integration: Introducing Integration by SubstitutionWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Recognise integrals that can be simplified using substitution
Choose an appropriate substitution $u = g (x)$ and rewrite $d x$ in terms of $d u$
Integrate fully in terms of $u$ , then substitute back
Apply substitution to definite integrals, changing the limits

Spotting a substitution

Substitution works well when the integral contains a function AND something proportional to its derivative multiplied alongside it — a strong hint that the chain rule was used to build the original expression.

1Let

u = x^{2} + 1

(the "inner" function), so

\frac{d u}{d x} = 2 x

, meaning

d u = 2 x d x

2Notice the integral already contains

2 x d x

exactly — replace it directly with

d u

\int u^{5} d u

3Integrate using the power rule:

\frac{u ^{6}}{6} + c

4Substitute back

u = x^{2} + 1

Answer

\frac{( x ^{2} + 1 ) ^{6}}{6} + c

Tip — A good candidate for $u$ is usually the expression INSIDE a bracket, root, or other composite structure — the part being "raised to a power" or "square-rooted", etc.

When du does not appear exactly

Often the exact multiple of the derivative is not sitting in the integral — you may need to rearrange to introduce a constant multiple, since constants can always be moved in and out of an integral freely.

1Let

u = x^{2} + 3

, so

d u = 2 x d x

, meaning

x d x = \frac{1}{2} d u

2Rewrite:

\int u^{4} \times \frac{1}{2} d u = \frac{1}{2} \int u^{4} d u

3Integrate:

\frac{1}{2} \times \frac{u ^{5}}{5} = \frac{u ^{5}}{10}

4Substitute back.

Answer

\frac{( x ^{2} + 3 ) ^{5}}{10} + c

Why this works

The missing factor of $\frac{1}{2}$ came directly from solving $d u = 2 x d x$ for $x d x$ — always solve explicitly for whatever appears in the original integral (here, $x d x$ ), not just for $d u$ itself.

Substitution in definite integrals

For a definite integral, you can either substitute back to $x$ before evaluating, OR — often faster — convert the LIMITS themselves into $u$ -values using the substitution, and evaluate directly in terms of $u$ without ever substituting back.

1Let

u = x^{2} + 1

d u = 2 x d x

, so

x d x = \frac{1}{2} d u

2Convert limits: when

x = 0

u = 1

; when

x = 2

u = 5

3Rewrite:

\int_{1}^{5} u^{3} \times \frac{1}{2} d u = \frac{1}{2} [\frac{u ^{4}}{4}]_{1}^{5} = \frac{1}{8} [u^{4}]_{1}^{5}

4Evaluate:

\frac{1}{8} (625 - 1) = \frac{1}{8} (624)

Answer

78

Tip — Converting the limits to $u$ -values avoids ever having to substitute back to $x$ — a real time-saver in definite integral questions.

Think like an examiner

Choose $u$ to be the "inner" function of the composite expression, then find $\frac{d u}{d x}$ and rearrange to match what appears in the original integral.
If the exact derivative is not present, look for a constant multiple — constants can always be pulled outside the integral sign.
For definite integrals, converting the limits into $u$ -values (rather than substituting back to $x$ at the end) is usually the more efficient method.

Common misconceptions

Forgetting to rewrite

d x

in terms of

d u

entirely (leaving a stray

x

in the integral).

Every instance of

x

(and

d x

) must be replaced — if a stray

x

remains after substitution, either the substitution is wrong or an extra step (solving for

x

) is needed.

Using the original x-limits after converting to a u-integral.

If you convert the limits to

u

-values, you must evaluate the

u

-integral using those NEW limits, not the original

x

-limits.

Forgetting to substitute back to x at the very end of an indefinite integral.

Unless you have converted the limits (only possible for definite integrals), your final indefinite-integral answer must be expressed back in terms of the original variable

x

Integration by substitution

u = g (x), d u = g \overset{ˊ}{(} x) d x

Define the substitution and its differential.

\int f (g (x)) g \overset{ˊ}{(} x) d x = \int f (u) d u

The substitution transforms the integral.

Stretch yourself

Use substitution to evaluate $\int_{1}^{2} \frac{x}{( x ^{2} + 1 ) ^{2}} d x$ , giving your answer as an exact fraction.

Hint — Let $u = x^{2} + 1$ , rewrite $x d x$ in terms of $d u$ , convert the limits, then integrate $u^{- 2}$ .

Questions students ask

How do I know when to use substitution?

Look for a composite function (something raised to a power, inside a root, etc.) multiplied by an expression that matches (or is a constant multiple of) the derivative of the "inner" part.

What if I cannot find the exact derivative in the integral?

Check for a constant multiple — you can always adjust by a constant factor, since constants move freely in and out of an integral. If a variable (non-constant) mismatch remains, substitution may not be the right technique.

Do I always need to substitute back to x?

For an indefinite integral, yes. For a definite integral, you can instead convert the limits into

u

-values and evaluate the whole thing in terms of

u

, skipping the substitute-back step entirely.

Key takeaways

Substitution reverses the chain rule: let $u$ be the inner function, rewrite $d x$ using $d u$ .
A constant multiple can always be introduced or removed to match what is in the integral.
For definite integrals, converting the limits to u-values avoids substituting back to x.
Always check every x has been replaced before integrating in terms of u.

How this fits the course

Build on

Integration by Parts

Leads to

Integration by Parts

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