Calculus Integration

Substitution rule

Q: What is u-substitution?

A method that reverses the chain rule: set u equal to the inner function, replace g prime of x dx with du, integrate in u, and substitute back.

Q: What do you do when the derivative is off by a constant?

Solve du equals g prime of x dx for dx and pull the constant outside the integral, then integrate normally.

Q: Do you substitute back at the end?

Yes, for an indefinite integral: replace u with g of x so the answer is in x. For a definite integral you may instead change the limits.

20 practice questions 0 video lessons Theory + worked examples

The Substitution Rule

California Calculus • Standard 17.0 • Integration

The Substitution Rule is a topic in Integration in the California Calculus Standards. It is aligned to Standard 17.0, which requires students to compute integrals using techniques of integration such as substitution.

The substitution rule is the reverse of the chain rule: setting \(u\) equal to an inner function turns \(\int f(g(x))\,g'(x)\,dx\) into the simpler \(\int f(u)\,du\).

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Practice questions

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Theory

Substitution (or u-substitution) is the reverse of the chain rule. When an integrand contains an inner function and its derivative, set \(u\) equal to the inner function; the integral collapses to a simple one in \(u\).

The chain rule differentiates \(f(g(x))\) as \(f'(g(x))g'(x)\). Substitution reads that backward:

\[\int f(g(x))\,g'(x)\,dx=\int f(u)\,du,\qquad u=g(x).\]

You choose \(u=g(x)\) (the inner function), compute \(du=g'(x)\,dx\), and rewrite the whole integral in terms of \(u\).

After integrating in \(u\), substitute back so the answer is in \(x\).

Key idea: look for an inner function whose derivative also appears (up to a constant). That derivative becomes part of \(du\).

Substitution reverses the chain rule: \(u=g(x)\), \(du=g'(x)\,dx\).

It finds areas under composite functions.

The substitution rule:

\[\int f(g(x))\,g'(x)\,dx=\int f(u)\,du\quad(u=g(x))\]

integral of f of g times g prime dx equals integral of f of u du

When a constant is missing: solve \(du=g'(x)\,dx\) for \(dx\) and carry the constant factor through the integral.

How to use substitution

Choose \(u\) to be the inner function.
Compute \(du=g'(x)\,dx\) and solve for \(dx\) if needed.
Rewrite the integral entirely in \(u\), integrate, then substitute back to \(x\).

Example 1 — Power form

Find \(\displaystyle\int 2x\,(x^2+1)^3\,dx\).

Solution

Let \(u=x^2+1\), so \(du=2x\,dx\).

\(\int 2x(x^2+1)^3\,dx\)	\(=\)	\(\int u^3\,du\)
	\(=\)	\(\dfrac{u^4}{4}+C=\dfrac{(x^2+1)^4}{4}+C\)

integral equals x squared plus one to the fourth over 4 plus C

Example 2 — A linear inside

Find \(\displaystyle\int (2x+1)^5\,dx\).

Solution

Let \(u=2x+1\), so \(dx=\dfrac{du}{2}\).

\(\int (2x+1)^5\,dx\)	\(=\)	\(\dfrac{1}{2}\int u^5\,du\)
	\(=\)	\(\dfrac{(2x+1)^6}{12}+C\)

integral equals 2x plus one to the sixth over 12 plus C

Example 3 — A trig inside

Find \(\displaystyle\int \cos(3x)\,dx\).

Solution

Let \(u=3x\), so \(dx=\dfrac{du}{3}\).

\(\int \cos(3x)\,dx\)	\(=\)	\(\dfrac{1}{3}\int \cos u\,du\)
	\(=\)	\(\dfrac{\sin(3x)}{3}+C\)

integral equals sin 3x over 3 plus C

Example 4 — An exponential

Find \(\displaystyle\int x\,e^{x^2}\,dx\).

Solution

Let \(u=x^2\), so \(x\,dx=\dfrac{du}{2}\).

\(\int x\,e^{x^2}\,dx\)	\(=\)	\(\dfrac{1}{2}\int e^{u}\,du\)
	\(=\)	\(\dfrac{e^{x^2}}{2}+C\)

integral equals e to the x squared over 2 plus C

Common pitfalls

Rewrite every \(x\). The integral must be entirely in \(u\) (including \(dx\)) before you integrate.

Handle a missing constant. If \(du\) needs a factor the integrand lacks, solve for \(dx\) and pull the constant out.

Substitute back. The final answer must be in the original variable \(x\), not \(u\).

Frequently asked questions

What is u-substitution?

A method that reverses the chain rule: set \(u\) equal to the inner function, replace \(g'(x)\,dx\) with \(du\), integrate in \(u\), and substitute back.

How do you choose u?

Pick the inner function whose derivative also appears in the integrand (possibly up to a constant factor).

What do you do when the derivative is off by a constant?

Solve \(du=g'(x)\,dx\) for \(dx\) and pull the constant outside the integral, then integrate normally.

Do you substitute back at the end?

Yes, for an indefinite integral: replace \(u\) with \(g(x)\) so the answer is in \(x\). (For a definite integral you may instead change the limits.)