What Is Integration by Substitution?
Integration by Substitution is a technique for evaluating integrals by changing the variable of integration from $x$ to a new variable $u = g(x)$. It is the integration counterpart of the chain rule, which is why it is often called u-substitution or the reverse chain rule. It is the first method most students reach for after the basic power, exponential, and trigonometric integration rules run out.
The method applies whenever the integrand can be written as a composite function multiplied by the derivative of its inner function:
$$\int f\big(g(x)\big),g'(x),dx ;=; \int f(u),du, \qquad \text{where } u = g(x),\ du = g'(x),dx.$$
The rule works because $g'(x),dx$ is exactly the piece that $du$ replaces. Once the integral is written purely in $u$, you integrate as usual and then swap $x$ back in at the end. Every indefinite answer carries a constant of integration $+C$, because differentiation destroys constants and substitution never recovers them.
How Do You Integrate By Substitution Step By Step?
The whole method is four short moves. The only judgement call is the first one: choosing $u$.
Choose $u = g(x)$ to be an inside function whose derivative $g'(x)$ also appears in the integrand (up to a constant factor).
Differentiate to get $du = g'(x),dx$, and solve for $dx$ if needed.
Substitute so that the integral is written entirely in $u$, with no stray $x$ left anywhere.
Integrate in $u$, then replace $u$ with $g(x)$ and add $+C$.
The choosing step is where readers stall. A reliable signal: scan the integrand for a chunk whose derivative is sitting right next to it. In $\int 2x(x^2+1)^5,dx$, the inside function $x^2+1$ has derivative $2x$, and that $2x$ is already in the integrand, so $u = x^2+1$ is the natural choice.
Geometrically, substitution is a change of scale on the horizontal axis. The factor $g'(x)$ measures how fast the $u$-axis stretches relative to the $x$-axis, and $du = g'(x),dx$ absorbs that stretch so the area under the curve is preserved. You are not changing the region, only the ruler you measure it with.
What Are Some Worked Examples Of Integration by Substitution?
Each example below is differentiated back to the original integrand, because the fastest way to catch a substitution slip is to check the answer.
Example 1: Evaluate $\int 2x(x^2+1)^5,dx$.
Let $u = x^2 + 1$, so $du = 2x,dx$. The integral becomes:
$$\int 2x(x^2+1)^5,dx = \int u^5,du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C.$$
Check by differentiating: $\dfrac{d}{dx}\left[\dfrac{(x^2+1)^6}{6}\right] = \dfrac{6(x^2+1)^5 \cdot 2x}{6} = 2x(x^2+1)^5$, the original integrand.
Final answer: $\dfrac{(x^2+1)^6}{6} + C$.
Example 2: Evaluate $\int \dfrac{\ln x}{x},dx$.
Let $u = \ln x$, so $du = \dfrac{1}{x},dx$. The $\dfrac{1}{x},dx$ in the integral is exactly $du$:
$$\int \frac{\ln x}{x},dx = \int u,du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C.$$
Check: $\dfrac{d}{dx}\left[\dfrac{(\ln x)^2}{2}\right] = \dfrac{2\ln x}{2}\cdot\dfrac{1}{x} = \dfrac{\ln x}{x}$. This pattern relies on the derivative of the logarithmic function $\ln x$.
Final answer: $\dfrac{(\ln x)^2}{2} + C$.
Example 3: Evaluate $\int x,e^{x^2},dx$.
Let $u = x^2$, so $du = 2x,dx$, which gives $x,dx = \dfrac{1}{2},du$:
$$\int x,e^{x^2},dx = \int e^{u}\cdot\frac{1}{2},du = \frac{1}{2}e^{u} + C = \frac{1}{2}e^{x^2} + C.$$
Check: $\dfrac{d}{dx}\left[\dfrac{1}{2}e^{x^2}\right] = \dfrac{1}{2}e^{x^2}\cdot 2x = x,e^{x^2}$.
Final answer: $\dfrac{1}{2}e^{x^2} + C$.
Example 4: Evaluate $\int \tan x,dx$.
Rewrite $\tan x = \dfrac{\sin x}{\cos x}$ and let $u = \cos x$, so $du = -\sin x,dx$, which gives $\sin x,dx = -,du$:
$$\int \tan x,dx = \int \frac{\sin x}{\cos x},dx = \int \frac{-,du}{u} = -\ln|u| + C = -\ln|\cos x| + C.$$
Check: $\dfrac{d}{dx}\left[-\ln|\cos x|\right] = -\dfrac{1}{\cos x}\cdot(-\sin x) = \dfrac{\sin x}{\cos x} = \tan x$. This is a standard result you will meet again in the differentiation of trigonometric functions.
Final answer: $-\ln|\cos x| + C$ (equivalently $\ln|\sec x| + C$).
Which Substitution Patterns Should You Recognise?
Most first-year substitutions fall into a handful of shapes. Learning to spot the pattern is faster than re-deriving each time.
Table: Standard integrands that Integration by Substitution solves, with the substitution to use.
Integrand pattern | Substitution | Result |
|---|---|---|
$\int \big(g(x)\big)^{n},g'(x),dx,\ n \neq -1$ | $u = g(x)$ | $\dfrac{\big(g(x)\big)^{n+1}}{n+1} + C$ |
$\int \dfrac{g'(x)}{g(x)},dx$ | $u = g(x)$ | $\ln\lvert g(x)\rvert + C$ |
$\int g'(x),e^{g(x)},dx$ | $u = g(x)$ | $e^{g(x)} + C$ |
$\int g'(x)\cos\big(g(x)\big),dx$ | $u = g(x)$ | $\sin\big(g(x)\big) + C$ |
$\int g'(x)\sin\big(g(x)\big),dx$ | $u = g(x)$ | $-\cos\big(g(x)\big) + C$ |
$\int \tan x,dx$ | $u = \cos x$ | $-\ln\lvert\cos x\rvert + C$ |
Every row is the same idea: the derivative of the inside function is present, so $du$ mops it up. The first row is the reverse power rule; the second is the reason $\int \frac{1}{x},dx = \ln|x| + C$ generalises so widely.
How Do You Use Integration by Substitution For A Definite Integral?
A definite integral has limits, and those limits are $x$-values. When you switch to $u$, the limits must switch too. If $u = g(x)$, then the bounds become $u = g(a)$ at the bottom and $u = g(b)$ at the top:
$$\int_{a}^{b} f\big(g(x)\big),g'(x),dx = \int_{g(a)}^{g(b)} f(u),du.$$
Example 5: Evaluate $\int_{0}^{1} 2x(x^2+1)^5,dx$.
Use $u = x^2 + 1$, $du = 2x,dx$, the same substitution as Example 1. Now convert the limits:
when $x = 0$, $u = 0^2 + 1 = 1$;
when $x = 1$, $u = 1^2 + 1 = 2$.
$$\int_{0}^{1} 2x(x^2+1)^5,dx = \int_{1}^{2} u^5,du = \left[\frac{u^6}{6}\right]_{1}^{2} = \frac{2^6 - 1^6}{6} = \frac{64 - 1}{6} = \frac{63}{6} = 10.5.$$
Once the limits are in $u$, you never go back to $x$: you evaluate in $u$ directly. The alternative is to find the antiderivative, return to $x$, and use the original $x$-limits $0$ and $1$. Both give $10.5$. What you must not do is change the limits to $u$ and then also back-substitute to $x$, which evaluates at the wrong numbers.
Geometrically, the area under $2x(x^2+1)^5$ from $x=0$ to $x=1$ equals the area under $u^5$ from $u=1$ to $u=2$. The substitution moved the interval, not the area.
Why Does Integration By Substitution Work?
Substitution is not a trick pulled from nowhere. It is the chain rule for derivatives read from right to left.
It undoes the chain rule. The chain rule says $\dfrac{d}{dx}\big[F(g(x))\big] = F'(g(x)),g'(x)$. If we call $F' = f$, then integrating both sides gives $\int f(g(x)),g'(x),dx = F(g(x)) + C$. Substitution is just this identity, organised so you can compute $F$ in the easy variable $u$.
The $du$ bookkeeping is exact. Writing $du = g'(x),dx$ is not sloppy shorthand; it is the statement that a thin slice $dx$ on the $x$-axis maps to a slice $du$ on the $u$-axis stretched by the factor $g'(x)$. Multiplying by $g'(x)$ keeps every slice, and therefore the whole area, unchanged.
The constant factor is free. If the derivative you need is present only up to a number, as in $\int x e^{x^2},dx$ where $du = 2x,dx$ but only $x,dx$ appears, you rescale with a constant. Constants pass straight through an integral, so this is always allowed.
That is the deep reason the method feels so natural once it clicks: differentiation and integration are inverse operations, and substitution is the chain-rule half of that inverse relationship made practical.
Who Invented Integration By Substitution?
The method grew out of the notation that made calculus usable, rather than from a single theorem.
Two later mathematicians turned the convenient notation into secure mathematics:
Leonhard Euler (1707–1783, Switzerland) used change of variables systematically across his integral calculus, treating substitution as a standard tool rather than a special case.
Augustin-Louis Cauchy (1789–1857, France) gave the definite integral its rigorous limit-of-sums definition, which is what justifies changing the limits when you change the variable.
Where Is Integration By Substitution Used In The Real World?
Any time a quantity accumulates and the rate depends on a changing inner variable, substitution tends to appear.
Kinematics and physics: finding distance from a velocity that depends on a changing quantity, or work done by a force that varies along a path, often needs a substitution to unwind the composite.
Pharmacokinetics: the total drug exposure in the body is the integral of a concentration that decays exponentially, exactly the $\int e^{g(t)}g'(t),dt$ shape substitution handles.
Probability and statistics: standardising a normal distribution, replacing $x$ with $z = \frac{x-\mu}{\sigma}$, is a substitution that turns a messy bell curve into the standard one whose areas are tabulated.
Signal processing and engineering: changing variables inside an integral is the routine first step when rescaling time or frequency in a transform.
Computer graphics and geometry: integrals for arc length, surface area, and volume are frequently set up as a composite and then reduced by substitution.
One method, learned on $\int 2x(x^2+1)^5,dx$, is the same one that measures drug exposure and standardises a bell curve. Mathematics reuses its good ideas everywhere the same shape appears.
What Are The Most Common Mistakes With Integration by Substitution?
These three errors account for most lost marks on substitution problems, and they match the exact questions readers ask on math forums and in class.
Forgetting to convert $dx$ into $du$.
Where it slips in:
A student picks $u = g(x)$, rewrites the composite part in $u$, but leaves the $dx$ and a stray $x$ untouched, then integrates a mix of $u$ and $x$.
Don't do this:
Do not integrate until every $x$ and the $dx$ are gone. An integral with both $u$ and $x$ in it cannot be evaluated.
The correct way:
Compute $du = g'(x),dx$ and use it to replace $dx$ (and any leftover $x$) completely. If a constant factor is missing, solve $du$ for $dx$ and carry the constant, as in $x,dx = \tfrac{1}{2},du$.
Not changing the limits on a definite integral (or changing them and back-substituting).
Where it slips in:
On $\int_{a}^{b}$, a student keeps the original $x$-limits while integrating in $u$, or converts the limits to $u$ and then also substitutes $x$ back before plugging the bounds in, double-counting the change.
Don't do this:
Do not evaluate $u$-limits on an $x$-antiderivative, and do not apply the change of variable twice.
The correct way:
Pick one lane. Either change the limits to $u = g(a)$ and $u = g(b)$ and evaluate entirely in $u$, or keep the answer in $x$ and use the original $x$-limits. Never both.
Choosing a $u$ whose derivative is not present.
Where it slips in:
A student sets $u$ to the most complicated-looking piece, but $g'(x)$ is nowhere in the integrand, so the $dx$ cannot be replaced and the integral does not simplify.
Don't do this:
Do not choose $u$ by appearance alone. If $g'(x)$ is not available (even up to a constant), substitution will stall.
The correct way:
Choose $u = g(x)$ so that $g'(x),dx$ is already sitting in the integrand. If it is not, substitution may be the wrong tool, and integration by parts or an algebraic rearrangement may be needed instead.
Practice Problems On Integration by Substitution
Work each one, then differentiate your answer back to check. Answers follow each line.
Find $\int 3x^2(x^3+2)^4,dx$.
(Answer: $u = x^3+2$, so $\dfrac{(x^3+2)^5}{5} + C$.)Find $\int \dfrac{2x}{x^2+1},dx$.
(Answer: $u = x^2+1$, so $\ln(x^2+1) + C$.)Find $\int \cos x,e^{\sin x},dx$.
(Answer: $u = \sin x$, so $e^{\sin x} + C$.)Find $\int (2x+1)(x^2+x)^3,dx$.
(Answer: $u = x^2+x$, so $\dfrac{(x^2+x)^4}{4} + C$.)Evaluate $\int_{0}^{2} x,e^{x^2},dx$.
(Answer: $u = x^2$, limits $0$ to $4$, so $\tfrac{1}{2}(e^4 - 1) \approx 26.7991$.)Find $\int \dfrac{(\ln x)^2}{x},dx$.
(Answer: $u = \ln x$, so $\dfrac{(\ln x)^3}{3} + C$.)
Where Should You Go Next After Integration by Substitution?
Substitution is one of two core integration techniques; the natural next steps build directly on it.
Integration of uv (by parts). The partner technique for products that substitution cannot reduce, such as $\int x e^{x},dx$.
The derivative formula sheet. Substitution is the chain rule reversed, so fluency with derivatives makes choosing $u$ automatic.
Calculus, the core ideas. Step back to see how differentiation and integration fit together as inverse operations.
If your child is moving into integral calculus, a live Bhanzu trainer teaches Integration by Substitution starting from the chain rule it reverses, in the Bhanzu math tutoring program. It is the same method that recurs in India's NCERT Class 12 Integrals chapter and under the US Common Core high-school standards, feeding straight into AP Calculus AB.
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