Integral of e^x: Antidifferentiate the natural exponential in one step

By Vegard Gjerde Based on Masterful Learning 9 min read
integral-of-e-to-the-x calculus integrals math learning-strategies

Integral of e^x gives the antiderivative of the natural exponential in one step: exdx=ex+C\int e^x\,dx = e^x + C. Its canonical condition is “always applies,” and the real fluency task is deciding when this rule is the direct pattern match versus when a larger structure means another rule governs the step first. That pattern check is a core habit in the Unisium Study System.

Unisium hero image titled Integral of e^x showing the principle equation and a conditions card.
The rule exdx=ex+C\int e^x\,dx = e^x + C for the direct pattern match exe^x.

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | FAQ | How This Fits


The Principle

The move: Integrate the exact local pattern exe^x in one step by copying the function and appending +C+C.

The invariant: This produces an antiderivative family whose derivative returns exe^x exactly.

Pattern: exdxex+C\int e^x\,dx \quad \longrightarrow \quad e^x + C

Applies directly ✓Does not apply directly ✗
exdxex+C\int e^x\,dx \to e^x + Ce2xdx↛e2x+C\int e^{2x}\,dx \not\to e^{2x} + C

Left: this step is a direct match for the rule. Right: the canonical condition still says “always applies,” but this guide is not the governing rule for the whole step because the exponent has extra structure.


Conditions of Applicability

Condition: always applies

Before applying, check: is this step a direct match for exdx\int e^x\,dx, or are you looking at a larger structure such as eg(x)e^{g(x)}, a sum, or a constant multiple?

  • The canonical condition is still “always applies.” The real decision is scope: this rule applies directly to the step exdx\int e^x\,dx, not automatically to every larger expression that contains base ee.
  • If the exponent is another function, such as in e2xdx\int e^{2x}\,dx, the issue is not a failed condition. The issue is that another method must handle the whole integral.
  • In mixed integrals, linearity rules such as the integral sum rule or integral constant multiple rule may govern the outer step while this rule applies to the exe^x term inside that larger chain.

Want the complete framework behind this guide? Read Masterful Learning.


Common Failure Modes

Failure mode: treat every exponential with base ee as though this rule governs the whole step -> write eg(x)dx=eg(x)+C\int e^{g(x)}\,dx = e^{g(x)} + C and lose the extra structure that another method must handle.

Debug: ask whether this is a direct exdx\int e^x\,dx match or a larger expression. If the exponent is anything other than the bare variable xx, stop and identify the governing rule before writing an antiderivative.


Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

  • Why does copying exe^x and adding +C+C produce a correct antiderivative family instead of only one antiderivative?
  • What does it mean to say that the derivative of the result returns the original integrand exactly?

For the Principle

  • How do you decide whether a step is a direct match for exdx\int e^x\,dx or a larger expression that needs a different outer rule first?
  • Why does “always applies” describe the canonical condition without making this the governing rule for every exponential integrand with base ee?

Between Principles

  • How does this rule connect to the derivative of e^x, and why do the derivative and integral versions both preserve the same functional form only for the exact pattern exe^x?

Generate an Example

  • Create one integral where this rule applies directly and one near-miss where it does not, then explain what structural check separates them.

Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Integrate the exact pattern e^x by copying the function and adding C.
Write the canonical equation: _____exdx=ex+C\int e^x\,dx = e^x + C
State the canonical condition: _____always applies

Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from (ex+x2)dx\int (e^x + x^2)\,dx, reach a simplified antiderivative while naming which local step uses the integral of exe^x rule.

StepExpressionOperation
0(ex+x2)dx\int (e^x + x^2)\,dx-
1exdx+x2dx\int e^x\,dx + \int x^2\,dxIntegral sum rule: split the sum into two integrals
2ex+x2dxe^x + \int x^2\,dxIntegral of exe^x rule on the exact local pattern exe^x
3ex+x33+Ce^x + \frac{x^3}{3} + CPower rule on x2x^2, then combine the constants of integration

Drills

Forward step (Format A)

Apply the rule to this direct-match integral.

exdx\int e^x\,dx

Reveal

This is a direct match for the rule, so the antiderivative is:

exdx=ex+C\int e^x\,dx = e^x + C


Apply the correct rule sequence and simplify.

5exdx\int -5e^x\,dx

Reveal

Use the integral constant multiple rule first, then apply the integral of exe^x rule to the exe^x term:

5exdx=5exdx=5ex+C\int -5e^x\,dx = -5\int e^x\,dx = -5e^x + C


Does this rule apply directly to the whole step? Explain before writing anything else.

e3xdx\int e^{3x}\,dx

Reveal

No. The canonical condition still says “always applies,” but this is not a direct match for the whole step.

The tempting move

e3xdx↛e3x+C\int e^{3x}\,dx \not\to e^{3x} + C

does not apply directly because it ignores the structure in the exponent. This guide covers the direct step exdx\int e^x\,dx, not every larger exponential integral.


Which items contain a direct or local use of the integral of exe^x rule in the solution chain?

(i) exdx\int e^x\,dx \quad (ii) ex2dx\int e^{x^2}\,dx \quad (iii) 4exdx\int 4e^x\,dx \quad (iv) 1xdx\int \frac{1}{x}\,dx

Reveal

(i) and (iii).

  • (i) is a direct match for the rule.
  • (iii) uses the integral of exe^x rule locally after factoring out the constant: 4exdx=4exdx\int 4e^x\,dx = 4\int e^x\,dx.
  • (ii) is a near-miss because the exponent is not the bare variable xx.
  • (iv) belongs to the integral of 1/x family, not to the exponential rule.

Action label (Format B)

What was done between these two steps?

exdxex+C\int e^x\,dx \quad \longrightarrow \quad e^x + C

Reveal

Integral of exe^x rule applied directly. The antiderivative copies the same exponential function and adds the constant of integration.


What was attempted between these two steps, and why does it not apply directly?

e2xdxe2x+C\int e^{2x}\,dx \quad \longrightarrow \quad e^{2x} + C

Reveal

An overextended direct use of the integral of exe^x rule was attempted.

The issue is not the canonical condition. The issue is scope: the whole step is not a direct exdx\int e^x\,dx match, so the move drops the extra structure in the exponent.


Name the rule combination used in this completed step.

(3ex+x4)dx3ex+x55+C\int (3e^x + x^4)\,dx \quad \longrightarrow \quad 3e^x + \frac{x^5}{5} + C

Reveal

The sum rule split the integral, the constant multiple rule pulled out 33, the integral of exe^x rule handled the exe^x term, and the power rule for integrals handled x4x^4.


Transition identification (Format C)

Which transition uses the integral of exe^x rule directly?

(x2+ex)dx(1)x2dx+exdx(2)x33+exdx(3)x33+ex+C\int (x^2 + e^x)\,dx \xrightarrow{(1)} \int x^2\,dx + \int e^x\,dx \xrightarrow{(2)} \frac{x^3}{3} + \int e^x\,dx \xrightarrow{(3)} \frac{x^3}{3} + e^x + C

Reveal

Transition (3) uses the integral of exe^x rule directly.

  • (1) is the sum rule.
  • (2) is the power rule on x2x^2.
  • (3) is the exact local step exdx=ex+C\int e^x\,dx = e^x + C.

Which transition does not apply directly, and why?

e2xdx(1)e2x+C(2)stop\int e^{2x}\,dx \xrightarrow{(1)} e^{2x} + C \xrightarrow{(2)} \text{stop}

Reveal

Transition (1) does not apply directly.

The move incorrectly treats e2xe^{2x} as though it were a direct exe^x match. This guide does not govern that whole-step rewrite.


Goal micro-chain (Format D)

Starting from (ex+x1)dx\int (e^x + x^{-1})\,dx, reach a correct final antiderivative by using the integral of exe^x rule only where it fits.

Reveal

First split the sum:

(ex+x1)dx=exdx+x1dx\int (e^x + x^{-1})\,dx = \int e^x\,dx + \int x^{-1}\,dx

Apply the integral of exe^x rule to the first term:

exdx=ex+C1\int e^x\,dx = e^x + C_1

Use the logarithm antiderivative on the second term:

x1dx=lnx+C2\int x^{-1}\,dx = \ln\left|x\right| + C_2

Combine constants:

(ex+x1)dx=ex+lnx+C\int (e^x + x^{-1})\,dx = e^x + \ln\left|x\right| + C


Starting from (2ex3x2)dx\int (2e^x - 3x^2)\,dx, reach a correct final antiderivative in the minimum number of moves.

Reveal

Split the integral and move constants out:

(2ex3x2)dx=2exdx3x2dx\int (2e^x - 3x^2)\,dx = 2\int e^x\,dx - 3\int x^2\,dx

Apply the local antiderivative rules:

2exdx3x2dx=2ex3(x33)+C2\int e^x\,dx - 3\int x^2\,dx = 2e^x - 3\left(\frac{x^3}{3}\right) + C

Simplify:

(2ex3x2)dx=2exx3+C\int (2e^x - 3x^2)\,dx = 2e^x - x^3 + C


Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Integrate 5ex+x35e^x + x^3 and simplify the result.

Full solution
StepExpressionMove
0(5ex+x3)dx\int (5e^x + x^3)\,dx-
15exdx+x3dx\int 5e^x\,dx + \int x^3\,dxIntegral sum rule
25exdx+x3dx5\int e^x\,dx + \int x^3\,dxIntegral constant multiple rule
35ex+x3dx5e^x + \int x^3\,dxIntegral of exe^x rule on the exact local pattern exe^x
45ex+x44+C5e^x + \frac{x^4}{4} + CPower rule on x3x^3, then combine constants

Check: differentiating 5ex+x44+C5e^x + \frac{x^4}{4} + C returns 5ex+x35e^x + x^3.


PrincipleRelationship
Indefinite integral as antiderivativeExplains why every correct antiderivative family ends with +C+C
Derivative of e^xGives the inverse local fact: exe^x keeps the same functional form under differentiation
Power rule for integralsHandles another direct antiderivative pattern often paired with exe^x in mixed integrals

FAQ

What is the integral of exe^x?

The integral of exe^x is ex+Ce^x + C. The function keeps the same form because differentiating exe^x returns exe^x again.

Why does the condition say “always applies” if I still have to be careful?

The canonical condition says there is no extra algebraic guard on the rule itself. The caution comes from scope: many larger expressions contain base ee without making this the governing rule for the whole step.

Can I use this rule on e2xdx\int e^{2x}\,dx?

Not as a one-step whole-rule move. The issue is not a failed condition. The issue is that this guide is not the governing rule for the whole integral when the exponent has extra structure.

Does the rule still work inside a longer integral like (ex+x5)dx\int (e^x + x^5)\,dx?

Yes. Use outer linearity rules such as the integral sum rule, then apply the integral of exe^x rule to the exe^x term inside that larger chain.

Why do I still add +C+C if the antiderivative looks obvious?

Indefinite integration returns a family of antiderivatives, not one single function. The constant accounts for every vertical shift with the same derivative.


How This Fits in Unisium

In Unisium, direct integration rules are trained as pattern recognition plus rule selection, not as isolated flashcards. This guide pairs the exact exe^x antiderivative fact with contrast cases so you can separate direct matches from near-misses, then stabilize that distinction with retrieval practice, self-explanation, and elaborative encoding. To extend the same workflow across your wider calculus stack, use Masterful Learning or practice directly in the Unisium app.

Explore further:

  • Integral sum rule - Split mixed integrals before choosing local antiderivative rules
  • Power rule for integrals - Pair the exe^x rule with another high-frequency direct antiderivative pattern
  • Integral of a^x - Compare the special base ee with the general constant-base exponential antiderivative
  • Derivative of e^x - Connect the antiderivative fact to its inverse derivative rule
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