U-substitution is the chain rule run backwards. The chain rule says $\frac{d}{dx}F(g(x)) = F'(g(x))\,g'(x)$. So whenever an integrand looks like $f(g(x))\,g'(x)$, an inner function times the derivative of that inner function, its antiderivative is $F(g(x)) + C$. Writing $u = g(x)$ and $du = g'(x)\,dx$ makes that pattern easy to see. OpenStax Calculus Volume 1 states it as $$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du = F(u) + C = F(g(x)) + C.$$
The five steps
OpenStax's problem-solving strategy:
- Choose an expression $g(x)$ in the integrand to call $u$, ideally one whose derivative $g'(x)$ is also there.
- Substitute $u = g(x)$ and $du = g'(x)\,dx$.
- Check that only $u$ is left. If not, go back and pick a different $u$.
- Integrate with respect to $u$.
- Write the answer in terms of $x$.
Worked example 1: the clean case
$\int 6x(3x^2+4)^4\,dx$. Let $u = 3x^2 + 4$, so $du = 6x\,dx$, which is already in the integrand: $$\int u^4\,du = \frac{u^5}{5} + C = \frac{(3x^2+4)^5}{5} + C.$$ Check by differentiating: the chain rule brings back $6x(3x^2+4)^4$.
Worked example 2: fixing a constant
$\int z\sqrt{z^2 - 5}\,dz$. Let $u = z^2 - 5$, so $du = 2z\,dz$. The integrand has only $z\,dz$, which is $\frac{1}{2}du$: $$\frac{1}{2}\int u^{1/2}\,du = \frac{1}{2}\cdot\frac{2}{3}u^{3/2} + C = \frac{1}{3}(z^2-5)^{3/2} + C.$$ You can move a constant factor in and out freely. You cannot do the same with a leftover variable.
Worked example 3: solving for $x$
$\int \frac{x}{\sqrt{x-1}}\,dx$. With $u = x - 1$, $du = dx$, but an $x$ is left over. Solve: $x = u + 1$. Then $$\int \frac{u+1}{\sqrt{u}}\,du = \int \left(u^{1/2} + u^{-1/2}\right)du = \frac{2}{3}u^{3/2} + 2u^{1/2} + C,$$ which OpenStax simplifies to $\frac{2}{3}(x-1)^{1/2}(x+2) + C$.
Definite integrals: change the limits
For a definite integral, change the limits to $u$ values: $\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$. OpenStax evaluates $\int_0^1 x^2(1+2x^3)^5\,dx$ with $u = 1 + 2x^3$, $du = 6x^2\,dx$. When $x = 0$, $u = 1$; when $x = 1$, $u = 3$. So $$\frac{1}{6}\int_1^3 u^5\,du = \frac{1}{36}\left(3^6 - 1^6\right) = \frac{182}{9}.$$ No substituting back is needed, because the limits already belong to $u$. This rests on the fundamental theorem of calculus.
Checkpoints
- $\int 3x^2(x^3 - 3)^2\,dx$
- $\int \sin t\cos^3 t\,dt$ (try $u = \cos t$)
- $\int_0^1 x e^{4x^2+3}\,dx$
- Is the substitution $u = 1 - x^2$ valid in $\int_0^2 x\sqrt{1-x^2}\,dx$? Why or why not?
Common mistakes
- Leaving some $x$ behind. Every $x$ and the $dx$ must be replaced before you integrate.
- Keeping the old limits. In a definite integral either change the limits to $u$ values or substitute back before evaluating, not neither.
- Dividing by a variable to fix $du$. Only constants can be adjusted that way.
- Forgetting $+C$ on an indefinite integral.
Related reading
- Chain rule intuition, the rule substitution reverses
- Integration by parts intuition and partial fractions intuition
- The fundamental theorem of calculus
- Calculus foundations topic map, all explainers and the sources page
Sources
Last reviewed: 12 November 2026.
Common questions
What is u-substitution?
It is a change of variables for integrals. You set u equal to an inner function g(x), replace g'(x) dx with du, integrate in u, then substitute back. It undoes the chain rule.
How do you choose u?
Pick an expression whose derivative also appears in the integrand, up to a constant factor. Often it is the inside of a power, a root, or the argument of a trig or exponential function.
What happens to the limits in a definite integral?
Change them to u values: the new limits are g(a) and g(b). Then you never need to substitute back.
What if du is off by a constant?
Multiply or divide by that constant. For example if du = 2z dz and the integral has z dz, use (1/2) du.