Calculus

U-Substitution: Finding the Inner Function

Substitution is the chain rule run backwards. The only genuinely difficult part is seeing which piece of the integrand is the inner function — everything after that is mechanical.

By the Gauth editorial teamHow this content is produced and checkedUpdated 10 September 2026

The idea

Differentiating a composite function produces a very specific shape. If $F$ is an antiderivative of $f$ , then the chain rule gives:

\frac{d}{d x} F (g (x)) = f (g (x)) \cdot g^{'} (x)

Reading that right to left is the whole technique:

\int f (g (x)) g^{'} (x) d x = F (g (x)) + C

So substitution is not really a method of integration. It is a method of recognition. Every integral that substitution solves was produced by the chain rule, and your job is to spot the fingerprints: an inner function, and its derivative loitering nearby as a factor.

Naming the inner function $u$ makes that pattern impossible to miss. With $u = g (x)$ and $d u = g^{'} (x) d x$ , the integral collapses to $\int f (u) d u$ — a problem in one letter, with no composition left in it. The substitution has not made the integral easier by magic; it has stripped away the layer that was hiding how easy it already was.

One consequence worth internalising early: the derivative has to be genuinely present. $\int 2 x (x^{2} + 1)^{5} d x$ works because $2 x$ is exactly $\frac{d}{d x} (x^{2} + 1)$ . Remove that factor and $\int (x^{2} + 1)^{5} d x$ is a different animal entirely — no substitution touches it, and you would have to expand the bracket.

When you need it

The signal is a composition paired with a derivative. Scan for these four shapes:

Bracket to a power, with the bracket's derivative outside: $\int x^{2} x^{3} + 1 d x$ , where $u = x^{3} + 1$ and $d u = 3 x^{2} d x$ .
A fraction whose numerator is the derivative of its denominator: $\int \frac{2 x}{x ^{2} + 1} d x$ . These always integrate to a logarithm.
An exponential or trig function with a non-trivial argument: $\int x e^{x^{2}} d x$ , $\int cos (5 x) d x$ .
Powers of a trig function multiplied by its derivative: $\int sin^{3} x cos x d x$ , where $u = sin x$ .

Substitution also produces several results that are worth recognising on sight rather than deriving each time. $\int tan x d x$ becomes $- ln ∣ cos x ∣ + C$ under $u = cos x$ , and $\int \frac{l n x}{x} d x$ becomes $\frac{1}{2} (ln x)^{2} + C$ under $u = ln x$ .

If the integrand is a product of two unrelated functions — a polynomial times an exponential, say, with no derivative relationship between them — substitution has nothing to grip and you want integration by parts instead. The two techniques cover very different territory, and trying the wrong one first is the most common way to waste ten minutes of an exam.

The method

1. Choose u as the inner function. The inside of the bracket, the thing under the root, the exponent, the argument of the trig function, the denominator. When two candidates compete, pick the one whose derivative you can actually find in the integrand.

2. Differentiate and check. Write $d u = g^{'} (x) d x$ and compare it with what is available. A constant mismatch is fixable — if $d u = 2 x d x$ and you only have $x d x$ , then $x d x = \frac{1}{2} d u$ . A mismatch by a factor of $x$ is not fixable, because constants can leave an integral and variables cannot.

3. Convert the limits straight away, if there are any. Push each $x$ -limit through $u = g (x)$ . Doing this at the moment you choose $u$ — rather than after the algebra — is what stops you from forgetting. It also means the new integral is entirely self-contained.

4. Rewrite the whole integral in u. Every $x$ and the $d x$ must go. If a stray $x$ refuses to convert, rearrange your substitution to express $x$ in terms of $u$ and replace it that way. A rewritten integral containing both letters is a statement that means nothing.

5. Integrate, then finish in the correct variable. Definite integral with converted limits: evaluate in $u$ and you are done — no substituting back. Indefinite integral: replace $u$ with the original expression and add $C$ . Then differentiate your answer mentally. Substitution answers are unusually quick to check, because the chain rule should hand back the original integrand in one line.

Changing the limits on a definite integral

0 of 6 steps

Try it yourself first — you'll remember it longer.

Worked examples

Worked example

\int 2 x (x^{2} + 1)^{5} d x

1
The bracket is raised to a power and its derivative, 2x, is sitting right there as a factor. Take the bracket as u.
$u = x^{2} + 1 d u = 2 x d x$
2
Replace the bracket by u and the whole of 2x dx by du. Nothing is left over.
$\int u^{5} d u$
3
Integrate the power in u.
$= \frac{u ^{6}}{6} + C$
4
Substitute back, since this integral is indefinite. Differentiating the result returns 6(x²+1)⁵·2x/6 = 2x(x²+1)⁵, which confirms it.
$= \frac{( x ^{2} + 1 ) ^{6}}{6} + C$

Answer

\int 2 x (x^{2} + 1)^{5} d x = \frac{( x ^{2} + 1 ) ^{6}}{6} + C

Worked example

\int x x - 1 d x

1
Take u as the expression under the root. Its derivative is 1, so du is just dx — but the stray x out front will not convert on its own.
$u = x - 1 d u = d x x = u + 1$
2
Rearranging the substitution gives x in terms of u, so the leftover factor can be replaced too.
$\int (u + 1) u d u$
3
Expand into separate powers of u. This is why the substitution works: the product becomes a sum.
$= \int (u^{3/2} + u^{1/2}) d u$
4
Integrate each power by adding one to the exponent and dividing.
$= \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C$
5
Substitute x − 1 back for u. Differentiating gives (x−1)^{3/2} + (x−1)^{1/2}, which factors to x√(x−1) as required.
$= \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C$

Answer

\int x x - 1 d x = \frac{2}{5} (x - 1)^{5/2} + \frac{2}{3} (x - 1)^{3/2} + C

Worked example

\int_{0}^{π /2} sin^{3} x cos x d x

1
The integrand is a power of sine multiplied by cosine, and cosine is the derivative of sine. That fixes the choice immediately.
$u = sin x d u = cos x d x$
2
Convert both limits through u = sin x before doing anything else.
$x = 0 \Rightarrow u = 0 x = \frac{π}{2} \Rightarrow u = 1$
3
Rewrite the integral. The cos x dx becomes du and the sin³x becomes u³.
$\int_{0}^{1} u^{3} d u$
4
Integrate and evaluate at the new limits. No substituting back is needed because the limits already travelled.
$[\frac{u ^{4}}{4}]_{0}^{1} = \frac{1}{4} - 0$

Answer

\int_{0}^{π /2} sin^{3} x cos x d x = \frac{1}{4}

Try one of your own

Type an integral and the substitution will be chosen and justified before it is applied, with the limits converted explicitly if the integral is definite. The integral calculator with steps covers substitution, parts, and partial fractions, and the calculus solver will take the whole question at once if it has several parts.

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Where people go wrong

Substituting without checking that du is there

Setting $u = x^{2} + 1$ in $\int (x^{2} + 1)^{5} d x$ and hoping. There is no $x$ factor, so $d x$ cannot be written in terms of $d u$ and the substitution stalls halfway. Always differentiate your candidate $u$ and physically look for that derivative in the integrand before you commit a single line to paper.

Pulling a variable outside the integral

Writing $\int x (x^{2} + 1)^{5} d x = x \int (x^{2} + 1)^{5} d x$ to manufacture the missing factor. Constants can move across an integral sign; variables cannot, because the integral is summing over the very quantity you tried to treat as fixed. The legitimate move on this integral is $x d x = \frac{1}{2} d u$ , which pulls out a one-half and leaves everything else intact.

Leaving an x behind

A rewritten integral containing both $u$ and $x$ is not a valid statement, and integrating it gives nonsense. Either express the stray $x$ in terms of $u$ , as in the second worked example above, or accept that the substitution was wrong. Scan the line before you integrate it; one stray letter is easy to see and expensive to miss.

Limits that did not travel

Changing the variable but keeping the original limits. In the example above, evaluating $ln ∣ u ∣$ at 1 and 3 instead of 2 and 10 gives $ln 3$ rather than $ln 5$ — a plausible looking answer that is simply wrong. Writing the new limits down at the moment you choose $u$ costs nothing and closes the gap. If you prefer to keep the old limits, you must substitute back to $x$ first, and then the original limits are correct again.

Practice

Two indefinite, one definite, one that needs you to spot a logarithm. Include the constant where the integral is indefinite.

Substitution practice

Answers are checked here — nothing is sent anywhere.

1
$\int (3 x + 2)^{4} d x$
2
$\int cos (5 x) d x$
3
$\int_{0}^{1} x (x^{2} + 3)^{2} d x$
4
$\int \frac{e ^{x}}{e ^{x} + 1} d x$

Frequently asked questions

How do I choose u?

Look for a function sitting inside another function — the inside of a bracket, the argument of a sine, the exponent of e, the thing under a root, or the denominator of a fraction. Take that as u, then check whether its derivative appears elsewhere in the integrand as a factor. If it does, up to a constant multiple, the substitution will work.

What if the derivative of u is off by a constant?

That is fine and very common. If u = x² + 1 then du = 2x dx, so an integrand containing only x dx means x dx = du/2. Pull the one-half out in front. What you cannot do is fix a mismatch involving the variable itself: if you are short an x, no constant will rescue the substitution.

Should I change the limits or substitute back at the end?

Both are valid, but changing the limits is safer and usually shorter. Once the limits are in terms of u you can finish entirely in u and never return to x. If you keep the original limits and substitute back, the risk is evaluating the antiderivative in u at the x-limits, which is a common and completely silent error.

What if there is a leftover x that will not convert?

Solve your substitution for x and replace it. For ∫x√(x−1) dx with u = x−1, the stray x becomes u+1, and the integral turns into ∫(u+1)√u du, which expands into two power terms. If solving for x is not possible either, the substitution is the wrong one.

How is this related to the chain rule?

It is the chain rule reversed. Differentiating F(g(x)) gives F'(g(x))·g'(x), so any integrand of that shape integrates straight back to F(g(x)). Substitution is just a bookkeeping device that makes that pattern visible by giving the inner function a name.

Keep going

Integral calculator with stepsEvery substitution shown explicitly, including the converted limits.The chain ruleThe forward direction. Substitution is far easier once this is automatic.Integration by partsWhat to reach for when the integrand is a product with no derivative relationship.Calculus solverDerivatives, integrals, limits, and series with the reasoning at each line.