The The Chain Rule produces derivatives like $\frac{d}{dx}(x^2 + 1)^6 = 6(x^2 + 1)^5\cdot 2x$. Substitution runs that rule backward: when an integrand looks like "outer function of $u$, times $u'$," we can integrate it. It extends the antiderivatives you can use with Evaluating Integrals with FTC Part 2.
Key idea
Substitution. Choose $u = g(x)$, usually an inside function, so that $du = g'(x)\,dx$ also appears (up to a constant factor). Rewrite the whole integral in $u$, integrate, then substitute back: $$\int f\big(g(x)\big)g'(x)\,dx = \int f(u)\,du.$$
Why it works
If $F' = f$, the chain rule says $\frac{d}{dx}F\big(g(x)\big) = f\big(g(x)\big)g'(x)$. So $F(g(x))$ is an antiderivative of $f(g(x))g'(x)$, and $F(u)$ is exactly $\int f(u)\,du$. The notation $du = g'(x)\,dx$ is a bookkeeping device that makes this automatic: it trades the factor $g'(x)\,dx$ for $du$.
| Integrand | Good choice of $u$ | $du$ |
|---|---|---|
| $2x(x^2 + 1)^5$ | $x^2 + 1$ | $2x\,dx$ |
| $x^2\sqrt{x^3 + 1}$ | $x^3 + 1$ | $3x^2\,dx$ |
| $\cos(3x)$ | $3x$ | $3\,dx$ |
| $\sin^4 x\cos x$ | $\sin x$ | $\cos x\,dx$ |
Worked example: The derivative is right there
Find $\displaystyle\int 2x(x^2 + 1)^5\,dx$.
- Let $u = x^2 + 1$; then $du = 2x\,dx$.
- The integral becomes $\int u^5\,du = \frac{u^6}{6} + C$.
- Back to $x$: $\frac{(x^2 + 1)^6}{6} + C$. Check: differentiating gives $(x^2 + 1)^5\cdot 2x$.
Answer: $\frac{(x^2 + 1)^6}{6} + C$
Worked example: Adjust by a constant
Find $\displaystyle\int x^2\sqrt{x^3 + 1}\,dx$.
- $u = x^3 + 1$, $du = 3x^2\,dx$, so $x^2\,dx = \frac{1}{3}du$.
- $\frac{1}{3}\int u^{1/2}\,du = \frac{1}{3}\cdot\frac{2}{3}u^{3/2} + C$.
Answer: $\frac{2}{9}(x^3 + 1)^{3/2} + C$
Worked example: Trig
Find $\displaystyle\int\cos(3x)\,dx$ and $\displaystyle\int\sin^4 x\cos x\,dx$.
- $u = 3x$, $dx = \frac{1}{3}du$: $\frac{1}{3}\int\cos u\,du = \frac{1}{3}\sin(3x) + C$.
- $u = \sin x$, $du = \cos x\,dx$: $\int u^4\,du = \frac{u^5}{5} = \frac{\sin^5 x}{5} + C$.
Answer: $\frac{1}{3}\sin(3x) + C$ and $\frac{\sin^5 x}{5} + C$
Choosing $u$
- Look for a function inside another function: inside a power, a root, a trig function, or an exponent.
- Check that its derivative appears as a factor, up to a constant. Constants can be fixed; missing variables cannot.
- If the first choice fails, try another. Substitution is a guess you check, not a recipe.
Substitution does not always work. $\int e^{x^2}\,dx$ has no $x$ factor to match $du = 2x\,dx$, and in fact $e^{x^2}$ has no antiderivative among the familiar functions. Compare $\int xe^{x^2}\,dx = \frac{1}{2}e^{x^2} + C$, where the extra $x$ makes everything work.
Common mistake
Leaving a mix of $x$ and $u$ in the integral, or ignoring the constant in $du$ (writing $\int x^2\sqrt{x^3 + 1}\,dx = \frac{2}{3}(x^3 + 1)^{3/2}$, which is 3 times too big). Before integrating, every $x$ and the $dx$ must be gone. Check by differentiating your answer.