Is ∫ e²ˣ dx just e²ˣ + C?

No — the chain rule strikes back. ∫ e²ˣ dx = e²ˣ/2 + C (via u-substitution with u = 2x). Differentiating e²ˣ gives 2e²ˣ, so the ÷2 is required.

Can I integrate x·eˣ with the exponential rule?

No — there is no product rule for integrals. ∫ x·eˣ dx is not eˣ·(x²/2); it needs integration by parts, a later technique.

Integral of eˣ Explained — Derivation, Examples & Quizzes

Q: What is the integral of eˣ?

∫ eˣ dx = eˣ + C. Since eˣ is its own derivative, it is also its own antiderivative — the function reproduces itself.

Q: Why is e special for integration?

e is defined as the base with d/dx[aˣ] = aˣ·ln a collapsing to exactly aˣ, because ln e = 1. For any other base a, ∫ aˣ dx = aˣ/ln a + C.

Q: What is ∫ 2ˣ dx?

∫ 2ˣ dx = 2ˣ/ln 2 + C. Differentiating 2ˣ gives an extra factor of ln 2, so the antiderivative must divide by ln 2 to cancel it.

Where it comes from

The problem is the reverse of a fact you already know: d/dx [e^x] = e^x. The function reproduces itself under differentiation — so running the film backwards, it must reproduce itself under integration too.

Why is e the special base? For a general base a, the derivative rule is d/dx [a^x] = a^x·ln a — an extra factor of ln a appears. The number e is defined as the base where that factor equals 1 (since ln e = 1), so the extra factor vanishes and e^x stands alone.

The tempting over-generalization:

Before reading on: d/dx [2^x] = 2^x·ln 2, so differentiation tacks on an extra ln 2. Running that backwards, what must you do to ∫ 2^x dx to cancel the extra factor? Take a guess before the reveal.

\int 2 x dx = 2 x + C ?? the tempting — and wrong — guess

Kill it by differentiating the claimed answer: d/dx [2^x + C] = 2^x·ln 2 ≠ 2^x. The derivative dragged in a factor of ln 2 ≈ 0.693, so the guess is only ~69% of what it should be. The repair — divide by ln 2:

d/dx [2^x/ln 2]

=

(2^x·ln 2)/ln 2 = 2^x

The ln 2 from the derivative cancels the ln 2 we divided by. So ∫ 2^x dx = 2^x/ln 2 + C — and for e^x, ln e = 1, so the division is invisible.

Intuition: e^x is the unique function (up to a constant multiple) that is its own rate of change — which is why it models unrestricted population growth and continuous compounding. Integrating it just hands you back the same growth curve, shifted by C.

Derivation

Again the guess-and-verify strategy: propose F(x) = e^x and differentiate to check. Then we generalize to any base a > 0.

F(x)

=

e^x

Step 1 — the guess. Since differentiating e^x reproduces it, e^x is the natural candidate for its own antiderivative.

F′(x)

=

e^x

Step 2 — verify. d/dx [e^x] = e^x, the defining property of e. So F′ = e^x: F is an antiderivative, and ∫ e^x dx = e^x + C. ∎

G(x)

=

a^x/ln a

Step 3 — generalize. For base a > 0, a ≠ 1, guess G(x) = a^x/ln a.

G′(x)

=

(a^x·ln a)/ln a = a^x

Step 4 — verify. The derivative contributes ×ln a, which cancels the ÷ln a. Hence ∫ a^x dx = a^x/ln a + C. Setting a = e recovers the hero formula, since ln e = 1.

Why +C is the whole story: any two antiderivatives of e^x differ by a constant (their difference has zero derivative), so e^x + C captures every one of them.

How to use it

Before reading on: e^x with exponent exactly x is the easy case. What do you think happens with e^2x — and which differentiation rule’s fingerprint should warn you that a plain e^x-style answer is wrong?

The procedure, every time:

Confirm the exponent is exactly x. The rule fires only for e^x alone (times constants). If the exponent is 2x, x², or −x, the chain rule left a fingerprint — see the traps below.
Write down e^x + C. That is genuinely the whole computation.
Other bases: ∫ a^x dx = a^x/ln a + C. Check by differentiating — the ln a cancels.
Constants factor out: ∫ 5e^x dx = 5e^x + C.

The two impostors

Impostor 1 — e^2x: ∫ e^2x dx is not e^2x + C. Differentiating e^2x gives 2e^2x (chain rule), so you must divide by 2: ∫ e^2x dx = e^2x/2 + C. This is u-substitution in disguise (u = 2x). Impostor 2 — x·e^x: there is no product rule for integrals — ∫ x·e^x dx ≠ e^x·(x²/2). That one needs integration by parts, a later technique. And ∫ e^x² dx has no elementary antiderivative at all.

Common mistake: writing ∫ e^3x dx = e^3x + C. Differentiate to check: 3e^3x ≠ e^3x. Whenever anything sits next to x in the exponent, expect a division — here ÷3.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: ∫ e^x dx

Recognize the self-reproducing function. The integrand is e^x with exponent exactly x.
Write the answer: e^x + C. (Why so short? Because d/dx [e^x] = e^x — the check is the proof.)

Common mistake: “doing something” to e^x — multiplying by x, adding an exponent. Resist the urge: the whole point of e is that nothing needs doing.

Your turn: Compute ∫ 2e^x dx.

Answer: 2e^x + C

The constant 2 factors out: 2·∫ e^x dx = 2e^x + C. Check: d/dx [2e^x] = 2e^x. ✓

Example 2 — term by term: ∫ (e^x + x²) dx

Split: ∫e^x dx + ∫x² dx.
First term: e^x (self-reproducing). Second term: x³/3 (power rule).
Combine with one +C: e^x + x³/3 + C.
Check: d/dx gives e^x + x². Matches ✓

Your turn: Compute ∫ (e^x + x⁴) dx.

Answer: e^x + x⁵/5 + C

Split: ∫ e^x dx + ∫ x⁴ dx = e^x + x⁵/5, one +C. Check: d/dx gives e^x + x⁴. ✓

Example 3 — definite: ∫₀¹ e^x dx

Antiderivative: e^x. (No +C needed for definite integrals — it would cancel.)
Evaluate: e¹ − e⁰ = e − 1 ≈ 1.718.
Sanity check: on [0,1], e^x runs from 1 to ≈2.718, so the area should be between 1·1 = 1 and 2.718·1 = 2.718. 1.718 sits inside. ✓

Common mistake: writing e¹ − e⁰ = e − 0. But e⁰ = 1, not 0 — anything to the power 0 is 1.

Your turn: Compute ∫₀² e^x dx.

Answer: e² − 1 ≈ 6.389

Antiderivative e^x; evaluate e² − e⁰ = e² − 1. Sanity: on [0,2], e^x runs 1→≈7.39, so the area sits between 2 and ≈14.8. 6.389 fits. ✓

Example 4 — another base: ∫ 3^x dx

Not base e, so expect a ÷ln 3 correction. Guess: 3^x/ln 3.
Verify: d/dx [3^x/ln 3] = (3^x·ln 3)/ln 3 = 3^x. Matches ✓
Answer: 3^x/ln 3 + C.

Common mistake: writing 3^x + C and skipping the ÷ln 3. The e^x rule is the special case ln e = 1 — every other base keeps its logarithm.

Your turn: Compute ∫ 5^x dx.

Answer: 5^x/ln 5 + C

Not base e, so pay the log tax: guess 5^x/ln 5. Verify: d/dx [5^x/ln 5] = (5^x·ln 5)/ln 5 = 5^x. ✓

Memorization tips

Say it aloud: “e to the x integrates to e to the x.” The sentence has no moving parts — that is the memory trick.
The ln a tax: every other base pays it — ∫ a^x dx = a^x/ln a + C. e is the base where the tax is 1, i.e. invisible.
The exponent test: before writing the answer, glance at the exponent. Exactly x? Fire the rule. Anything else (2x, −x, x²)? Stop — chain-rule fingerprints need u-substitution.
Pair with its derivative: d/dx [e^x] = e^x and ∫ e^x dx = e^x + C are the same fact read in two directions. Learn one, get the other free.
No product rule: x·e^x is not e^x·(x²/2). If you catch yourself “integrating each factor,” stop.

Final challenge

Five mixed questions — bases, traps, and a rewrite. Score 5/5 and the exponential is yours.

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the integral of e^x?

∫ e^x dx = e^x + C. Since e^x is its own derivative, it is also its own antiderivative — the function reproduces itself.

Why is e special for integration?

e is defined as the base with d/dx [a^x] = a^x·ln a collapsing to exactly a^x, because ln e = 1. For any other base a, ∫ a^x dx = a^x/ln a + C.

What is ∫ 2^x dx?

∫ 2^x dx = 2^x/ln 2 + C. Differentiating 2^x gives an extra factor of ln 2, so the antiderivative must divide by ln 2 to cancel it.

Is ∫ e^2x dx just e^2x + C?

No — the chain rule strikes back. ∫ e^2x dx = e^2x/2 + C (via u-substitution with u = 2x). Differentiating e^2x gives 2e^2x, so the ÷2 is required.

Can I integrate x·e^x with the exponential rule?

No — there is no product rule for integrals. ∫ x·e^x dx is not e^x·(x²/2); it needs integration by parts, a later technique.

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The exponential integral

Where it comes from

Derivation

How to use it

The two impostors

Worked examples

Example 1 — the basic move: ∫ ex dx

Example 2 — term by term: ∫ (ex + x²) dx

Example 3 — definite: ∫01 ex dx

Example 4 — another base: ∫ 3x dx