U-Substitution: Identifying Composite Functions and Derivative Factors

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Hello! In the previous lesson, you found antiderivatives by rewriting integrands into familiar forms. That approach works when algebra exposes powers, logarithms, or other standard integrals. But an expression such as $cos (x^{2} + 1)$ has a function inside another function, so integrating the outer function alone would miss something.

This lesson introduces substitution, a way to recognize and reverse the chain rule. You will learn how to choose an inner expression, check whether its derivative is present, and carry an integral through to an answer in the original variable.

The chain rule in reverse

Recall what happens when you differentiate a composite function:

\frac{d}{d x} sin (x^{2} + 1) = cos (x^{2} + 1) \cdot 2 x .

The factor $2 x$ comes from differentiating the inner expression $x^{2} + 1$ . Reading this calculation backward gives

\int 2 x cos (x^{2} + 1) d x = sin (x^{2} + 1) + C .

More generally, if $F^{'} (u) = f (u)$ , then the chain rule gives

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

So the pattern to look for is

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

The expression $g (x)$ is the inner expression. Its derivative $g^{'} (x)$ is the derivative factor. Spotting both is more reliable than simply choosing the longest expression in an integrand.

Watch how this pattern becomes a method. The first example shows an exact derivative match; the second shows why a numerical mismatch does not prevent substitution.

Integration Using The Substitution Rule

Watch “Integration Using The Substitution Rule” by Professor Dave Explains to see how the inner expression, its derivative, and the final differentiation check fit together.

In the first example, follow the choice of u, the conversion to du, and the return to x. Then watch the chain-rule connection: notice that a composite expression alone is not enough; its derivative factor must also be accounted for. In the second example, focus on how dividing by a constant repairs an almost-matching derivative.

Making—and completing—a substitution

Consider

\int 6 x (3 x^{2} + 4)^{4} d x .

The power $(3 x^{2} + 4)^{4}$ suggests choosing its base as the new variable. Check that choice by differentiating it:

u = 3 x^{2} + 4, d u = 6 x d x .

The rest of the integrand is exactly $6 x d x$ . Replacing both parts changes the entire integral into one variable:

\int 6 x (3 x^{2} + 4)^{4} d x = \int u^{4} d u = \frac{u ^{5}}{5} + C = \frac{( 3 x ^{2} + 4 ) ^{5}}{5} + C .

That last replacement matters: because the original question was in $x$ , an indefinite-integral answer should be returned to $x$ . Differentiating it checks both the power-rule step and the choice of substitution:

\frac{d}{d x} (\frac{( 3 x ^{2} + 4 ) ^{5}}{5}) = (3 x^{2} + 4)^{4} (6 x) .

The notation $d u = g^{'} (x) d x$ is a convenient way to keep track of the derivative factor when $u = g (x)$ . The chain rule is what justifies the method. It does not permit replacing $d x$ with $d u$ while leaving unrelated $x$ -terms unchanged.

A dependable sequence is to choose an inner expression, calculate its derivative, account for every factor in the integrand, integrate entirely in $u$ , and substitute back. If you still have an unexplained $x$ in the new integral, the change of variable is not finished.

When the derivative is present only up to a constant

An exact match is convenient, but a constant multiple works just as well. For example,

\int x x^{2} + 9 d x

contains the inner expression $x^{2} + 9$ , whose derivative is $2 x$ . The integrand has $x$ , not $2 x$ , so write

u = x^{2} + 9, d u = 2 x d x, x d x = \frac{1}{2} d u .

Now the square root is $u$ , and the whole integral is expressed in $u$ :

\int x x^{2} + 9 d x = \frac{1}{2} \int u^{1/2} d u = \frac{1}{3} u^{3/2} + C = \frac{1}{3} (x^{2} + 9)^{3/2} + C .

The factor $\frac{1}{2}$ is essential. Without it, differentiating the proposed answer would give twice the original integrand.

This same reasoning applies when the derivative is a constant and no obvious multiplier is written. For instance, with $u = 3 x - 1$ , we have $d u = 3 d x$ , so

\int sin (3 x - 1) d x = \frac{1}{3} \int sin u d u = - \frac{1}{3} cos (3 x - 1) + C .

In both examples, the adjustment is a constant that can be moved outside the integral. You cannot generally insert a missing factor of $x$ , or another variable expression, and compensate by pulling out a constant.

Recognizing other familiar forms after substitution

Substitution does not supply a new table of antiderivatives. It changes the variable so that the forms from the previous lesson become usable. A particularly important one is $\int 1/ u d u = ln ∣ u ∣ + C$ .

Look at

\int \frac{2 x}{x ^{2} - 9} d x .

Choosing the denominator as $u$ works because its derivative is the numerator times $d x$ :

u = x^{2} - 9, d u = 2 x d x .

Therefore,

\int \frac{2 x}{x ^{2} - 9} d x = \int \frac{1}{u} d u = ln ∣ u ∣ + C = ln ∣ x^{2} - 9∣ + C .

The absolute value allows the answer on intervals where $x^{2} - 9$ is negative as well as where it is positive. Work on an interval that does not contain $x = 3$ or $x = - 3$ , where the original integrand is undefined.

An exponential works similarly. If $u = 5 x - 2$ , then $d u = 5 d x$ , giving

\int e^{5 x - 2} d x = \frac{1}{5} \int e^{u} d u = \frac{1}{5} e^{5 x - 2} + C .

These examples have different outer functions, but the decision is the same each time: differentiate the proposed inner expression and compare the result with what remains in the integrand.

Deciding whether a choice of $u$ actually helps

It is easy to see two expressions in an integrand and assume that one must be the derivative of the other. The direction matters: the derivative factor must come from the expression inside the outer function. The following short reading concentrates on mistakes that reveal why this test is necessary.

𝘶-substitution (article)

Read the common-mistakes portion of Khan Academy’s “u-substitution” article to sharpen your choice of the inner expression and derivative factor.

Find the subsection “Common mistake: getting incorrect expressions for u or du.” Begin with three pitfalls and continue through the subsections “Common mistake: not realizing u-substitution is called for” and “Another common mistake: confusing the inner function and its derivative.” Stop after the example involving x^2\cos(2x), before “Sometimes we need to multiply/divide the integral by a constant.” Focus on which expression is inside which function, rather than just whether two derivatives appear somewhere on the page.

For example, $\int x^{2} cos (2 x) d x$ does not offer a direct substitution of the kind used above. If $u = 2 x$ , then $d u = 2 d x$ , but the factor $x^{2}$ remains. You could rewrite $x^{2}$ in terms of $u$ , yet the resulting integral would not be a standard form from the previous lesson. The presence of a composite function does not guarantee that substitution alone will evaluate an integral.

A leftover $x$ is not always a dead end, however. If it can be rewritten entirely in terms of $u$ and the resulting integral is familiar, the substitution may still help. For example, in

\int (x + 2) x + 1 d x,

take $u = x + 1$ . Then $d u = d x$ and $x + 2 = u + 1$ , so

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

The test is not merely “Did I find a $u$ ?” It is “After changing variables, can I write every part of the integral in $u$ and integrate what remains?”

Key takeaways

Substitution reverses the chain rule. Choose an inner expression $u = g (x)$ , calculate $d u = g^{'} (x) d x$ , and check whether the integrand contains that derivative factor—possibly differing by a constant. Then rewrite the whole integral in $u$ , use a familiar antiderivative, return to $x$ , and differentiate to check your answer.

When the needed derivative factor is not available and rewriting in $u$ does not simplify the integral, do not force the method. The next lesson introduces integration by parts, which is useful for certain products that substitution does not resolve.