AP Calculus AB and BC

Integral of e^(3x): Answer, Steps, and Mistakes

Q: Is e^(3x)/3 the same as (1/3)e^(3x)?

Yes. \frac{e^{3x}}{3} and \frac{1}{3}e^{3x} are two ways of writing the same expression, and both are correct answers for \int e^{3x}\,dx.

Q: How do you integrate e^(kx) in general?

\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C for any nonzero constant k. The substitution u = kx gives dx = \frac{du}{k}, which is where the \frac{1}{k} comes from.

The integral of e^(3x) is e^(3x)/3 + C, the same as (1/3)e^(3x) + C. Because the exponent 3x has slope 3, you divide by that inner coefficient. Differentiating e^(3x)/3 gives (1/3) times 3 times e^(3x), which is e^(3x).

\int e^{3x}\,dx = \frac{e^{3x}}{3} + C

The substitution, step by step

The integrand is a composite: the natural exponential wrapped around $3x$ . That composite structure is the cue for u-substitution, with the inside function as $u$ .

u = 3x \quad\Longrightarrow\quad \frac{du}{dx} = 3 \quad\Longrightarrow\quad dx = \frac{du}{3}

Replace the exponent and the differential, then pull the constant $\frac{1}{3}$ outside the integral.

\int e^{3x}\,dx = \int e^{u}\cdot\frac{du}{3} = \frac{1}{3}\int e^{u}\,du = \frac{1}{3}e^{u} + C

Substituting $u = 3x$ back gives the answer in the original variable.

\int e^{3x}\,dx = \frac{e^{3x}}{3} + C

The general pattern

$\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ for any nonzero constant $k$ . A linear exponent always divides by its own coefficient.

Why the factor of 1/3 appears

Differentiating the answer shows exactly where the $\frac{1}{3}$ earns its place. The chain rule attaches a factor of $3$ to the derivative of $e^{3x}$ .

\frac{d}{dx}\left(e^{3x}\right) = e^{3x}\cdot\frac{d}{dx}(3x) = 3e^{3x}

So $e^{3x}$ by itself differentiates to three times too much. The $\frac{1}{3}$ out front cancels that stray factor.

\frac{d}{dx}\left(\frac{e^{3x}}{3} + C\right) = \frac{1}{3}\cdot 3e^{3x} = e^{3x}

Common mistakes

Multiplying by $3$ instead of dividing. Because differentiating $e^{3x}$ multiplies by $3$ , integrating must divide by $3$ . The answer is $\frac{e^{3x}}{3}$ , not $3e^{3x}$ .
Leaving off the coefficient and writing $e^{3x} + C$ , which differentiates to $3e^{3x}$ , not $e^{3x}$ .
Applying the power rule. The variable sits in the exponent, not the base, so $\int e^{3x}\,dx$ is not $\frac{e^{3x+1}}{3x+1}$ . Exponentials integrate by dividing by the inner coefficient.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of e^(3x) from 0 to 1?

$\int_0^1 e^{3x}\,dx = \left[\frac{e^{3x}}{3}\right]_0^1 = \frac{e^3 - 1}{3}$ . The $\frac{1}{3}$ stays through the evaluation.

Is e^(3x)/3 the same as (1/3)e^(3x)?

Yes. $\frac{e^{3x}}{3}$ and $\frac{1}{3}e^{3x}$ are two ways of writing the same expression, and both are correct answers for $\int e^{3x}\,dx$ .

How do you integrate e^(kx) in general?

$\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ for any nonzero constant $k$ . The substitution $u = kx$ gives $dx = \frac{du}{k}$ , which is where the $\frac{1}{k}$ comes from.