The Central Idea
The chain rule says
\[ \frac{d}{dx}F(g(x)) = F'(g(x))g'(x). \]Substitution reverses that structure:
\[ \boxed{ \int F'(g(x))g'(x)\,dx = F(g(x))+C }. \]The central recognition pattern is
\[ \boxed{ \text{outer function of an inner expression} \times \text{derivative of the inner expression} }. \]Why Substitution Works
Suppose
\[ u=g(x). \]Differentiate both sides:
\[ \frac{du}{dx}=g'(x). \]In differential notation,
\[ \boxed{du=g'(x)\,dx}. \]Therefore,
\[ \int F'(g(x))g'(x)\,dx \]becomes
\[ \int F'(u)\,du. \]The complicated composite expression has been replaced by a simpler single-variable integral.
Example: The Simplest Substitution Pattern
Evaluate
\[ \int 2x(x^2+1)^5\,dx. \]The inner expression is
\[ x^2+1. \]Its derivative is
\[ 2x. \]That derivative factor is already present in the integrand.
Carrying Out the Substitution
Let
\[ u=x^2+1. \]Then
\[ du=2x\,dx. \]The integral becomes
\[ \int u^5\,du. \]Apply the power rule:
\[ \frac{u^6}{6}+C. \]Substitute back:
\[ \boxed{ \frac{(x^2+1)^6}{6}+C }. \]Checking by Differentiation
Differentiate
\[ \frac{(x^2+1)^6}{6}. \]By the chain rule,
\[ \frac16 \cdot 6(x^2+1)^5 \cdot 2x. \]Simplifying,
\[ 2x(x^2+1)^5. \]The original integrand returns, so the antiderivative is correct.
What Should We Choose for \(u\)?
A good substitution usually replaces the expression that is nested inside another operation.
Common candidates include:
- the expression inside a power;
- the expression inside a radical;
- the expression inside an exponential;
- the expression inside a logarithm;
- the expression inside a trigonometric function;
- the denominator of a rational expression;
- any recurring composite expression whose derivative also appears nearby.
The Inner-Function Heuristic
In
\[ (3x+4)^7, \]the inner expression is
\[ 3x+4. \]In
\[ e^{x^2}, \]the inner expression is
\[ x^2. \]In
\[ \sin(5x^3), \]the inner expression is
\[ 5x^3. \]Substitution often begins by naming that inner expression \(u\).
The Derivative May Match Only Up to a Constant
Evaluate
\[ \int x(x^2+4)^3\,dx. \]Let
\[ u=x^2+4. \]Then
\[ du=2x\,dx. \]But the integral contains only \(x\,dx\).
Solve for that piece:
\[ \boxed{x\,dx=\frac12\,du}. \]Compensating for the Constant
The integral becomes
\[ \frac12\int u^3\,du. \]Therefore,
\[ \frac12\cdot\frac{u^4}{4}+C = \frac{u^4}{8}+C. \]Substitute back:
\[ \boxed{ \frac{(x^2+4)^4}{8}+C }. \]Example: Linear Inner Function
Evaluate
\[ \int (5x-2)^8\,dx. \]Let
\[ u=5x-2. \]Then
\[ du=5\,dx, \]so
\[ dx=\frac15\,du. \]Therefore,
\[ \int (5x-2)^8\,dx = \frac15\int u^8\,du. \]Thus,
\[ \boxed{ \frac{(5x-2)^9}{45}+C }. \]Example: Exponential Substitution
Evaluate
\[ \int 6x e^{3x^2}\,dx. \]Let
\[ u=3x^2. \]Then
\[ du=6x\,dx. \]Therefore,
\[ \int e^u\,du=e^u+C. \]Substitute back:
\[ \boxed{ e^{3x^2}+C }. \]Example: Trigonometric Substitution Pattern
Evaluate
\[ \int 4x^3\cos(x^4)\,dx. \]Let
\[ u=x^4. \]Then
\[ du=4x^3\,dx. \]The integral becomes
\[ \int \cos u\,du. \]Therefore,
\[ \boxed{ \sin(x^4)+C }. \]Example: Sine with a Linear Inner Function
Evaluate
\[ \int \sin(7x+1)\,dx. \]Let
\[ u=7x+1. \]Then
\[ du=7\,dx, \]so
\[ dx=\frac17\,du. \]Thus,
\[ \frac17\int\sin u\,du = -\frac17\cos u+C. \]Therefore,
\[ \boxed{ -\frac17\cos(7x+1)+C }. \]The Logarithmic Substitution Pattern
Recall
\[ \int \frac1u\,du=\ln|u|+C. \]Therefore,
\[ \boxed{ \int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)|+C }. \]This is one of the most useful substitution patterns.
Example: Rational Logarithmic Pattern
Evaluate
\[ \int \frac{2x}{x^2+5}\,dx. \]Let
\[ u=x^2+5. \]Then
\[ du=2x\,dx. \]So
\[ \int\frac1u\,du = \ln|u|+C. \]Since \(x^2+5>0\),
\[ \boxed{ \ln(x^2+5)+C }. \]Example: Logarithmic Pattern with a Constant Adjustment
Evaluate
\[ \int \frac{x}{x^2-9}\,dx. \]Let
\[ u=x^2-9. \]Then
\[ du=2x\,dx, \]so
\[ x\,dx=\frac12du. \]Therefore,
\[ \frac12\int\frac1u\,du = \frac12\ln|u|+C. \]Hence,
\[ \boxed{ \frac12\ln|x^2-9|+C }. \]Example: Radical Substitution
Evaluate
\[ \int \frac{3x^2}{\sqrt{x^3+4}}\,dx. \]Let
\[ u=x^3+4. \]Then
\[ du=3x^2\,dx. \]Rewrite the radical:
\[ \frac1{\sqrt{u}}=u^{-1/2}. \]Thus,
\[ \int u^{-1/2}\,du = 2u^{1/2}+C. \]Therefore,
\[ \boxed{ 2\sqrt{x^3+4}+C }. \]Example: A Nested Radical
Evaluate
\[ \int x\sqrt{x^2+1}\,dx. \]Let
\[ u=x^2+1, \]so
\[ x\,dx=\frac12du. \]Then
\[ \frac12\int u^{1/2}\,du = \frac12\cdot\frac{u^{3/2}}{3/2}+C. \]Therefore,
\[ \boxed{ \frac13(x^2+1)^{3/2}+C }. \]Sometimes Algebra Comes Before Substitution
The needed derivative factor may be hidden until the integrand is simplified.
For example,
\[ \int \frac{x^3+x}{x^2+1}\,dx. \]Factor the numerator:
\[ x^3+x=x(x^2+1). \]Therefore,
\[ \frac{x^3+x}{x^2+1}=x. \]The integral reduces immediately to
\[ \boxed{\frac{x^2}{2}+C}. \]Always simplify before choosing a complicated method.
Do Not Expand a Useful Composite Structure
Consider
\[ \int 2x(x^2+1)^{10}\,dx. \]Expanding
\[ (x^2+1)^{10} \]would create a long polynomial.
Substitution preserves the useful structure:
\[ u=x^2+1, \qquad du=2x\,dx. \]Then
\[ \boxed{ \int u^{10}\,du = \frac{u^{11}}{11}+C }. \]Replace the Entire Integral Consistently
After choosing \(u\), the goal is to rewrite the integral entirely in terms of \(u\) and \(du\).
For example, after setting
\[ u=x^2+1, \]we should not leave an unmatched \(x\) inside the transformed integral.
A clean substitution has the form
\[ \boxed{ \int \text{expression in }x\,dx \longrightarrow \int \text{expression in }u\,du }. \]How to Recognize a Poor Substitution
Suppose we try
\[ u=x^2+1 \]in
\[ \int (x^2+1)^5\,dx. \]Then
\[ du=2x\,dx, \]but no factor of \(x\) appears in the integral.
The substitution does not simplify the problem.
Substitution is most useful when the derivative of the chosen inner expression is present, at least up to a constant factor.
Substitution in Definite Integrals
Substitution also works for definite integrals.
There are two valid workflows:
- substitute, integrate in \(u\), return to \(x\), then use the original bounds;
- substitute and immediately convert the bounds from \(x\)-values to \(u\)-values.
The second method is often cleaner because no back-substitution is needed.
Changing the Bounds
Suppose
\[ u=g(x) \]and the original definite integral runs from
\[ x=a \]to
\[ x=b. \]Then the new bounds are
\[ \boxed{ u=g(a) \qquad\text{and}\qquad u=g(b) }. \]Example: Definite Integral with Changed Bounds
Evaluate
\[ \int_0^1 2x(x^2+1)^3\,dx. \]Let
\[ u=x^2+1. \]Then
\[ du=2x\,dx. \]Change the bounds:
\[ x=0\longrightarrow u=1, \] \[ x=1\longrightarrow u=2. \]Completing the Definite Integral
The integral becomes
\[ \int_1^2u^3\,du. \]Therefore,
\[ \left[\frac{u^4}{4}\right]_1^2. \]Evaluate:
\[ \frac{16}{4}-\frac14. \]Thus,
\[ \boxed{\frac{15}{4}}. \]Alternative Definite-Integral Workflow
Using the same example,
\[ \int_0^1 2x(x^2+1)^3\,dx, \]we can first find the antiderivative:
\[ \frac{(x^2+1)^4}{4}. \]Then evaluate using the original \(x\)-bounds:
\[ \left[ \frac{(x^2+1)^4}{4} \right]_0^1. \]This produces the same answer:
\[ \boxed{\frac{15}{4}}. \]Do Not Mix Variables and Bounds
Once an integral has been rewritten entirely in \(u\), its bounds must also be \(u\)-values.
For example, this is inconsistent:
\[ \int_0^1 u^3\,du. \]if \(0\) and \(1\) are still the original \(x\)-bounds.
Either:
- keep the original bounds and return to \(x\) before evaluating; or
- convert the bounds and remain in \(u\).
Example: Definite Logarithmic Substitution
Evaluate
\[ \int_0^1 \frac{2x}{x^2+1}\,dx. \]Let
\[ u=x^2+1, \qquad du=2x\,dx. \]The bounds become
\[ x=0\to u=1, \qquad x=1\to u=2. \]Therefore,
\[ \int_1^2\frac1u\,du = [\ln u]_1^2. \]Hence,
\[ \boxed{\ln2}. \]Example: Definite Trigonometric Substitution
Evaluate
\[ \int_0^{\sqrt{\pi}}2x\cos(x^2)\,dx. \]Let
\[ u=x^2, \qquad du=2x\,dx. \]The bounds become
\[ 0\to0, \qquad \sqrt{\pi}\to\pi. \]Therefore,
\[ \int_0^\pi\cos u\,du = [\sin u]_0^\pi. \]Thus,
\[ \boxed{0}. \]Differential Notation as a Matching Tool
If
\[ u=g(x), \]then
\[ du=g'(x)\,dx. \]This notation helps us identify what factor must accompany the inner function.
For
\[ u=x^3+2, \]we get
\[ du=3x^2\,dx. \]So an integrand containing \(x^2\,dx\) matches after multiplying by \(1/3\).
Solving the Differential Relation
Sometimes it is convenient to solve for \(dx\).
If
\[ u=4x-7, \]then
\[ du=4\,dx. \]Therefore,
\[ \boxed{dx=\frac14\,du}. \]This is especially useful when the derivative of the inner function is a nonzero constant.
Think in Terms of Function Composition
An expression such as
\[ (1+x^5)^{12} \]has the structure
\[ F(g(x)), \]where
\[ g(x)=1+x^5. \]If the integrand also includes
\[ 5x^4, \]then the chain-rule pattern is complete.
This compositional viewpoint is more reliable than memorizing isolated substitution tricks.
A General Power Pattern
If \(n\neq-1\), then
\[ \boxed{ \int [g(x)]^n g'(x)\,dx = \frac{[g(x)]^{n+1}}{n+1}+C }. \]This is simply the ordinary power rule applied after
\[ u=g(x). \]A General Exponential Pattern
Since
\[ \int e^u\,du=e^u+C, \]substitution gives
\[ \boxed{ \int e^{g(x)}g'(x)\,dx = e^{g(x)}+C }. \]General Trigonometric Patterns
Substitution gives
\[ \boxed{ \int \cos(g(x))g'(x)\,dx = \sin(g(x))+C }, \]and
\[ \boxed{ \int \sin(g(x))g'(x)\,dx = -\cos(g(x))+C }. \]These are reverse-chain-rule versions of familiar trigonometric derivatives.
More Reverse-Chain-Rule Patterns
Similarly,
\[ \boxed{ \int \sec^2(g(x))g'(x)\,dx = \tan(g(x))+C }, \] \[ \boxed{ \int \sec(g(x))\tan(g(x))g'(x)\,dx = \sec(g(x))+C }. \]The important feature is always the same: the derivative of the inner expression accompanies the outer derivative pattern.
Example: A Multi-Step Algebraic Substitution
Evaluate
\[ \int \frac{6x^2}{(x^3+1)^2}\,dx. \]Let
\[ u=x^3+1. \]Then
\[ du=3x^2\,dx. \]Since the numerator contains \(6x^2\,dx\),
\[ 6x^2\,dx=2\,du. \]The integral becomes
\[ 2\int u^{-2}\,du. \]Therefore,
\[ -2u^{-1}+C. \]Hence,
\[ \boxed{ -\frac{2}{x^3+1}+C }. \]Example: Fractional Power after Substitution
Evaluate
\[ \int \frac{x}{(x^2+4)^{2/3}}\,dx. \]Let
\[ u=x^2+4. \]Then
\[ x\,dx=\frac12du. \]Therefore,
\[ \frac12\int u^{-2/3}\,du. \]Apply the power rule:
\[ \frac12\cdot\frac{u^{1/3}}{1/3}+C. \]Thus,
\[ \boxed{ \frac32(x^2+4)^{1/3}+C }. \]Common Substitution Structures
| Integrand structure | Natural substitution |
|---|---|
| \([g(x)]^n g'(x)\) | \(u=g(x)\) |
| \(e^{g(x)}g'(x)\) | \(u=g(x)\) |
| \(\cos(g(x))g'(x)\) | \(u=g(x)\) |
| \(\sin(g(x))g'(x)\) | \(u=g(x)\) |
| \(\dfrac{g'(x)}{g(x)}\) | \(u=g(x)\) |
| \(\dfrac{g'(x)}{\sqrt{g(x)}}\) | \(u=g(x)\) |
Substitution Does Not Solve Every Integral
Some integrals do not contain a useful reverse-chain-rule pattern.
For example,
\[ \int x e^x\,dx \]does not simplify by choosing \(u=x\) or \(u=e^x\).
That integral is better suited to a later method: integration by parts.
Choosing the right method is part of integration.
Common Error: Choosing \(u\) Without Checking Its Derivative
A substitution is useful only if it simplifies the integral.
After choosing \(u\), always compute
\[ du. \]Then compare \(du\) with the remaining factors in the integrand.
If the pieces do not match, the substitution may be poor or incomplete.
Common Error: Ignoring a Missing Constant Factor
If
\[ u=x^2+1, \]then
\[ du=2x\,dx. \]Therefore,
\[ x\,dx=\frac12du. \]Replacing \(x\,dx\) directly by \(du\) would make the answer twice as large as it should be.
Common Error: Forgetting to Substitute Back
For an indefinite integral, the final answer should normally be written in the original variable.
If
\[ u=x^2+4 \]and the transformed antiderivative is
\[ \frac{u^5}{5}+C, \]then the final answer is
\[ \boxed{ \frac{(x^2+4)^5}{5}+C }. \]Common Error: Mixing \(x\) and \(u\)
An expression such as
\[ \int u^3x\,du \]usually indicates an incomplete substitution.
Once the substitution is made, rewrite every required part consistently in the new variable whenever possible.
Common Error: Keeping Old Bounds after Changing Variables
If a definite integral has been rewritten in \(u\), then its limits should be \(u\)-values.
Convert
\[ x=a,\quad x=b \]into
\[ u=g(a),\quad u=g(b). \]Do not combine \(u\)-notation with unconverted \(x\)-bounds.
Common Error: Adding \(+C\) to a Definite Integral
Substitution does not change the distinction between definite and indefinite integrals.
An indefinite integral ends with
\[ +C. \]A definite integral evaluates to one number and does not.
Common Error: Expanding Before Looking for Structure
If an integrand already has the form
\[ g'(x)[g(x)]^n, \]expanding the power usually destroys the obvious substitution pattern.
Preserve useful composition when possible.
A Reliable Procedure for Indefinite \(u\)-Substitution
- Inspect the integrand for a composite inner expression.
- Choose \(u\) equal to that inner expression.
- Compute \(du\).
- Match the derivative factor, adjusting by constants when necessary.
- Rewrite the entire integral in \(u\).
- Integrate using familiar rules.
- Substitute back into the original variable.
- Add \(+C\).
- Differentiate the result to verify it.
A Reliable Procedure for Definite \(u\)-Substitution
- Choose \(u=g(x)\).
- Compute \(du=g'(x)\,dx\).
- Rewrite the integrand in \(u\).
- Convert the lower and upper bounds using \(u=g(x)\).
- Evaluate the transformed definite integral.
- Do not substitute back to \(x\) if the bounds were converted.
- Do not add \(+C\).
A Fast Recognition Checklist
Before integrating, ask:
- Is there a function inside another function?
- Does the derivative of the inner function appear elsewhere?
- Does it match exactly or only up to a constant?
- Would replacing the inner expression simplify the integrand?
- Can I rewrite the entire problem in one new variable?
If the answers line up, substitution is likely the right method.
Example: Full Substitution Workflow
Evaluate
\[ \int 15x^2(5x^3-1)^4\,dx. \]Choose
\[ u=5x^3-1. \]Then
\[ du=15x^2\,dx. \]The integral becomes
\[ \int u^4\,du. \]Therefore,
\[ \frac{u^5}{5}+C. \]Substitute back:
\[ \boxed{ \frac{(5x^3-1)^5}{5}+C }. \]Verifying the Full Workflow
Differentiate
\[ \frac15(5x^3-1)^5. \]By the chain rule,
\[ \frac15 \cdot 5(5x^3-1)^4 \cdot 15x^2. \]Simplifying,
\[ 15x^2(5x^3-1)^4. \]The original integrand returns.
Example: Full Definite-Integral Workflow
Evaluate
\[ \int_1^2 3x^2(x^3+2)^2\,dx. \]Let
\[ u=x^3+2, \qquad du=3x^2\,dx. \]Convert the bounds:
\[ x=1\to u=3, \] \[ x=2\to u=10. \]Therefore,
\[ \int_3^{10}u^2\,du = \left[\frac{u^3}{3}\right]_3^{10}. \]Hence,
\[ \boxed{ \frac{1000-27}{3} = \frac{973}{3} }. \]Practice 1: Direct Power Substitution
Evaluate:
\[ \int 6x(x^2+3)^4\,dx. \]Practice 2: Constant Adjustment
Evaluate:
\[ \int x(x^2+5)^6\,dx. \]Practice 3: Linear Inner Function
Evaluate:
\[ \int (4x+1)^5\,dx. \]Practice 4: Exponential Pattern
Evaluate:
\[ \int 8x e^{4x^2}\,dx. \]Practice 5: Trigonometric Pattern
Evaluate:
\[ \int 3x^2\cos(x^3)\,dx. \]Practice 6: Logarithmic Pattern
Evaluate:
\[ \int \frac{4x}{2x^2+7}\,dx. \]Practice 7: Radical Pattern
Evaluate:
\[ \int \frac{2x}{\sqrt{x^2+9}}\,dx. \]Practice 8: Definite Integral
Evaluate:
\[ \int_0^1 2x(x^2+2)^2\,dx. \]Practice 9: Definite Logarithmic Integral
Evaluate:
\[ \int_0^2 \frac{2x}{x^2+1}\,dx. \]Practice 10: Choose the Substitution
For
\[ \int x^4\sin(x^5+2)\,dx, \]identify a useful choice of \(u\).
Practice 11: Identify the Missing Factor
If
\[ u=x^3+1, \]what factor must accompany \(dx\) in order to match \(du\)?
Practice 12: Identify the Error
A student sets
\[ u=x^2+1 \]in
\[ \int x(x^2+1)^3\,dx \]and writes
\[ \int u^3\,du. \]Explain the missing factor.
Practice 13: Bounds after Substitution
If
\[ u=x^2+4 \]and the original bounds are
\[ x=1 \quad\text{to}\quad x=3, \]what are the corresponding \(u\)-bounds?
Practice 14: Method Recognition
Would substitution be a natural first method for
\[ \int x e^x\,dx? \]Explain briefly.
Practice Answers
Practice 1
Let
\[ u=x^2+3, \qquad du=2x\,dx. \]Then
\[ 6x\,dx=3\,du. \]Therefore,
\[ 3\int u^4\,du = \frac35u^5+C. \]Hence,
\[ \boxed{ \frac35(x^2+3)^5+C }. \]Practice 2
Let
\[ u=x^2+5. \]Then
\[ x\,dx=\frac12du. \]So
\[ \frac12\int u^6\,du = \frac{u^7}{14}+C. \]Therefore,
\[ \boxed{ \frac{(x^2+5)^7}{14}+C }. \]Practice 3
Let
\[ u=4x+1. \]Then
\[ dx=\frac14du. \]Thus,
\[ \frac14\int u^5\,du = \frac{u^6}{24}+C. \]Hence,
\[ \boxed{ \frac{(4x+1)^6}{24}+C }. \]Practice 4
Let
\[ u=4x^2. \]Then
\[ du=8x\,dx. \]Therefore,
\[ \boxed{ e^{4x^2}+C }. \]Practice 5
Let
\[ u=x^3, \qquad du=3x^2\,dx. \]Then
\[ \int\cos u\,du=\sin u+C. \]Therefore,
\[ \boxed{ \sin(x^3)+C }. \]Practice 6
Let
\[ u=2x^2+7. \]Then
\[ du=4x\,dx. \]Therefore,
\[ \boxed{ \ln(2x^2+7)+C }. \]Practice 7
Let
\[ u=x^2+9, \qquad du=2x\,dx. \]Then
\[ \int u^{-1/2}\,du = 2u^{1/2}+C. \]Therefore,
\[ \boxed{ 2\sqrt{x^2+9}+C }. \]Practice 8
Let
\[ u=x^2+2, \qquad du=2x\,dx. \]The bounds become
\[ 0\to2, \qquad 1\to3. \]Thus,
\[ \int_2^3u^2\,du = \left[\frac{u^3}{3}\right]_2^3. \]Therefore,
\[ \boxed{ \frac{27-8}{3} = \frac{19}{3} }. \]Practice 9
Let
\[ u=x^2+1, \qquad du=2x\,dx. \]The bounds become
\[ 0\to1, \qquad 2\to5. \]Therefore,
\[ \int_1^5\frac1u\,du = [\ln u]_1^5. \]Hence,
\[ \boxed{\ln5}. \]Practice 10
The inner expression is
\[ x^5+2. \]Since its derivative contains \(x^4\), choose
\[ \boxed{u=x^5+2}. \]Practice 11
If
\[ u=x^3+1, \]then
\[ \boxed{du=3x^2\,dx}. \]The matching factor is \(3x^2\).
Practice 12
Since
\[ du=2x\,dx, \]we have
\[ x\,dx=\frac12du. \]Therefore, the transformed integral should be
\[ \boxed{ \frac12\int u^3\,du }. \]Practice 13
When \(x=1\),
\[ u=1^2+4=5. \]When \(x=3\),
\[ u=3^2+4=13. \]Therefore, the new bounds are
\[ \boxed{5\text{ to }13}. \]Practice 14
No. The integrand
\[ x e^x \]does not contain a useful inner-function-and-derivative pattern.
It is better suited to
\[ \boxed{\text{integration by parts}}. \]What to Remember
Substitution reverses the chain rule.
If
\[ u=g(x), \]then
\[ \boxed{du=g'(x)\,dx}. \]The central pattern is
\[ \boxed{ \int F'(g(x))g'(x)\,dx = F(g(x))+C }. \]For definite integrals, bounds can be changed directly:
\[ x=a\to u=g(a), \qquad x=b\to u=g(b). \]The most important habit is:
look for a nested expression and then check whether its derivative is also present.
Where We Go Next
Substitution reverses the chain rule. The next lesson, Integration by Parts, reverses the product rule.
That method becomes especially useful for products such as
\[ x e^x, \qquad x\sin x, \qquad x\ln x, \]where ordinary substitution does not simplify the integral.