Integrating Using Substitution
Introduced| English |
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| u-substitution/juː ˌsʌbstɪˈtjuːʃn/ |
The chain rule, run in reverse
- The chain rule made derivatives of composite functions. Integration must undo it.
- u-substitution 换元法 (or u-sub) is the reverse of the chain rule.
- It spots a composite function together with its inner derivative and simplifies by renaming.
- Master it and a huge class of otherwise-impossible integrals becomes routine.
u-substitution is the reverse of which differentiation rule?
It undoes the chain rule.
Rename the inner function as $u$
- Look for an inner function $g(x)$ whose derivative $g'(x)$ also appears (up to a constant) in the integrand.
- Let $u=g(x)$; then $du=g'(x)\,dx$.
- Rewrite the whole integral in terms of $u$ — the messy $x$-expression turns into a simple $u$-integral.
- Then integrate in $u$ and substitute back $u=g(x)$ at the end.
A composite ready for u-sub
y = ax³ + bx
An integrand like $2x(x^2+1)^3$ hides an inner function and its derivative — the signal to substitute.
For $\int 2x(x^2+1)^3\,dx$, a good choice is $u=$
$u=x^2+1$ has $du=2x\,dx$, which is present.
A clean example of the pattern
- $\displaystyle\int 2x\,(x^2+1)^3\,dx$. Inner function $u=x^2+1$, so $du=2x\,dx$.
- The $2x\,dx$ in the integral is exactly $du$ — a perfect match.
- The integral becomes $\displaystyle\int u^3\,du=\dfrac{u^4}{4}+C$.
- Substitute back: $\dfrac{(x^2+1)^4}{4}+C$.
$\int 2x(x^2+1)^3\,dx$ equals...
$\int u^3\,du=\tfrac{u^4}{4}+C$, then back-substitute.
Handle the constant, or the limits
- If $du$ is off by a constant factor, just multiply/divide to fix it (e.g. $x\,dx=\tfrac12\,du$).
- For a definite integral, either substitute back to $x$ before evaluating, or change the limits to $u$-values.
- Changing limits is cleaner: convert $a,b$ into $u(a),u(b)$ and never return to $x$.
- Always end an indefinite answer in terms of the original variable $x$.
Evaluate $\int_0^1 3x^2(x^3+1)^2\,dx$ (let $u=x^3+1$). Enter a decimal.
$\int_1^2 u^2\,du=\tfrac{8-1}{3}=\tfrac73\approx2.333$.
A correct u-substitution converts $dx$ entirely into $du$, leaving no stray $x$.
A leftover $x$ or $dx$ is the classic mistake.
u-substitution works when the integrand contains the inner function's...
The inner derivative $g'(x)$ must be present (times a constant).
u-sub works only when the inner derivative is present (up to a constant). If $g'(x)$ isn't there, you can't just invent it. And don't forget to convert $dx$ into $du$ — leaving a stray $dx$ or a leftover $x$ is the classic error. For definite integrals, either change the limits or back-substitute, not a mix.
Evaluate $\displaystyle\int_0^1 3x^2\,(x^3+1)^2\,dx$.
- Let $u=x^3+1$, so $du=3x^2\,dx$. Limits: $x=0\Rightarrow u=1$, $x=1\Rightarrow u=2$.
- Integral becomes $\displaystyle\int_1^2 u^2\,du=\Big[\tfrac{u^3}{3}\Big]_1^2=\tfrac{8}{3}-\tfrac{1}{3}=\tfrac{7}{3}$.
u-substitution reverses the chain rule: set $u=g(x)$ (the inner function), replace $g'(x)\,dx$ with $du$, integrate in $u$, then substitute back (or change the limits for a definite integral). It works when the inner derivative appears up to a constant factor. Always convert $dx\to du$ fully.