U-Substitution in Integration

U-substitution is the standard integration method for expressions like $\int f(g(x))g'(x)\,dx$. You choose the inside expression as $u$, replace the matching derivative part with $du$, and turn the integral into something simpler.

Use it when one function is clearly nested inside another and the derivative of the inner expression is also present, exactly or up to a nonzero constant factor.

What U-Substitution Means

The pattern is:

$$\int f(g(x))g'(x)\,dx$$

If you let $u = g(x)$, then $du = g'(x)\,dx$, so the integral becomes

$$\int f(u)\,du$$

That is the whole idea. A messy inner expression becomes a single variable, so the antiderivative is easier to recognize.

How To Spot When U-Substitution Works

U-substitution works best when the integrand has a clear composite structure. In plain language, one function is sitting inside another, and some version of the inner derivative is also present.

Common patterns include powers such as $(x^2+1)^5$, radicals such as $\sqrt{3x-2}$, exponentials such as $e^{x^2}$, and trig expressions such as $\cos(x^3)$.

If the derivative of the inside expression is missing entirely, substitution may not help. If it is only off by a nonzero constant factor, you can often fix that by factoring the constant in or out first.

Worked Example: $\int \frac{x}{x^2+1}\,dx$

Find

$$\int \frac{x}{x^2+1}\,dx$$

The denominator has an inside expression $x^2+1$, and its derivative is $2x$. The numerator is only half of that, which is close enough for substitution.

Let

$$u = x^2 + 1$$

Then

$$du = 2x\,dx$$

so

$$x\,dx = \frac{1}{2}du$$

Rewrite the integral:

$$\int \frac{x}{x^2+1}\,dx = \int \frac{1}{u}\cdot \frac{1}{2}\,du = \frac{1}{2}\int \frac{1}{u}\,du$$

Now integrate:

$$\frac{1}{2}\int \frac{1}{u}\,du = \frac{1}{2}\ln|u| + C$$

Substitute back:

$$\int \frac{x}{x^2+1}\,dx = \frac{1}{2}\ln(x^2+1) + C$$

Since $x^2+1 > 0$ for all real $x$, writing $\ln(x^2+1)$ is fine here.

Why U-Substitution Makes Sense

Differentiation with the chain rule says an outer function picks up a factor from the inner derivative. U-substitution runs that idea backward. It groups the inner expression into one symbol and treats the derivative piece as the matching differential.

That is why the method is not random pattern matching. It is a structured undoing of the chain rule.

Common U-Substitution Mistakes

1. Choosing $u$ without checking whether its derivative also appears. If the matching derivative is not there, the substitution may not simplify anything.
2. Forgetting the constant factor adjustment. In the example above, using $du = 2x\,dx$ but ignoring the $\frac{1}{2}$ gives the wrong answer.
3. Mixing variables after substituting. Once you rewrite in terms of $u$, the integral should stay entirely in $u$ until you substitute back.
4. Forgetting $+C$ on an indefinite integral.
5. Keeping the variable as $u$ in a definite integral but still using the old $x$-limits. If you integrate in $u$, the bounds must also change to $u$-values.

U-Substitution With Definite Integrals

For a definite integral, you can handle the last step in either of two correct ways.

One option is to substitute back to $x$ and use the original limits. The other option is to keep the answer in $u$ and change the bounds immediately.

For example, if

$$\int_0^1 2x\cos(x^2)\,dx$$

and you let $u=x^2$, then the new bounds are $u=0$ and $u=1$, so

$$\int_0^1 2x\cos(x^2)\,dx = \int_0^1 \cos u\,du = \sin 1$$

The important condition is consistency: do not mix $u$ with $x$-bounds.

Where U-Substitution Is Used

U-substitution is one of the first major integration techniques in calculus because many antiderivatives are not direct formula matches until you rewrite them.

It appears in basic calculus courses, differential equations, probability, physics, and engineering whenever a quantity is naturally built from an inner expression and its rate of change.

Try A Similar U-Substitution Problem

Try

$$\int (3x^2)\,e^{x^3}\,dx$$

before looking anything up. If you choose $u=x^3$, the integral should collapse quickly. After you finish, check whether your final answer is back in $x$ and whether you kept the constant factor correctly.