# Integral of e^(2x): Formula, Steps and Examples

> The integral of e^(2x) is ½e^(2x) + C, and in general ∫e^(kx) dx = (1/k)e^(kx) + C. See the substitution, a graph and worked examples.

Source: https://practicecalculus.com/integrals/e-to-the-2x
Updated: 2026-11-06

$$ \int e^{2x}\,dx = \frac{1}{2}e^{2x} + C $$

## At a glance

- **Antiderivative:** $\frac{1}{2}e^{2x} + C$
- **General form:** $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$ for $k \ne 0$
- **Method:** Substitution $u = 2x$, $du = 2\,dx$
- **From 0 to 1:** $\frac{e^2 - 1}{2} \approx 3.195$

## Key takeaways

- The integral of $e^{2x}$ is $\frac{1}{2}e^{2x} + C$.
- For any constant $k \ne 0$, $\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$. Divide by the coefficient of $x$.
- This only works when the exponent is linear. For $e^{x^2}$ there is no such shortcut.
- You can also see it from $e^{2x} = (e^x)^2$, but the substitution is faster.

The **integral of $e^{2x}$ is $\dfrac{1}{2}e^{2x} + C$**. More generally, to integrate $e^{kx}$ you keep the exponential and divide by $k$.

$$\int e^{2x}\,dx = \frac{1}{2}e^{2x} + C \qquad\qquad \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C\quad (k \ne 0)$$

## Graph of e²ˣ and its antiderivative

```graph
title: 'y = e^(2x) and its antiderivative ½e^(2x)'
caption: 'The shaded area from $0$ to $1$ is $\frac{e^2 - 1}{2} \approx 3.195$. The dashed antiderivative is always half the height of $e^{2x}$, since $\frac{1}{2}e^{2x}$ is just $e^{2x}$ scaled down.'
x: [-2, 1.2]
y: [-0.5, 10]
functions:
  - expr: e^(2x)
    label: y = e^(2x)
  - expr: 0.5*e^(2x)
    label: y = ½e^(2x) (antiderivative)
    dashed: true
shade:
  expr: e^(2x)
  from: 0
  to: 1
points:
  - [0, 1, '(0, 1)']
```

## How to integrate e²ˣ step by step

Use [u-substitution](/rules/u-substitution-rule).

1. Let $u = 2x$. Then $du = 2\,dx$, so $dx = \frac{1}{2}\,du$.
2. Rewrite: $\displaystyle\int e^{2x}\,dx = \int e^u\cdot\frac{1}{2}\,du = \frac{1}{2}\int e^u\,du$.
3. Integrate: $\frac{1}{2}e^u + C$, using the [integral of eˣ](/integrals/e-to-the-x).
4. Substitute back: $\frac{1}{2}e^{2x} + C$.

The same steps with $u = kx$ give $\frac{1}{k}e^{kx} + C$. With a constant added, $u = kx + b$, you get $\frac{1}{k}e^{kx + b} + C$.

**Another way to see it.** The rule "divide by the inside slope" works for every linear inside: $\int\cos(3x)\,dx = \frac{1}{3}\sin 3x$, $\int (2x+1)^5\,dx = \frac{1}{12}(2x+1)^6$. It is the reverse of the chain rule multiplying by the inside slope.

## Check by differentiating

$$\frac{d}{dx}\left(\frac{1}{k}e^{kx}\right) = \frac{1}{k}\cdot e^{kx}\cdot k = e^{kx}$$

The [chain rule](/rules/chain-rule) brings out a factor $k$, and the $\frac{1}{k}$ cancels it. See the [derivative of e²ˣ](/derivatives/e-to-the-2x) for the forward direction.

## Definite integrals

| Integral | Work | Value |
| --- | --- | --- |
| $\int_0^1 e^{2x}\,dx$ | $\frac{1}{2}(e^2 - 1)$ | $\approx 3.195$ |
| $\int_0^{\ln 3} e^{2x}\,dx$ | $\frac{1}{2}(9 - 1)$ | $4$ |
| $\int_{-\infty}^{0} e^{2x}\,dx$ | $\frac{1}{2}(1 - 0)$ | $\frac{1}{2}$ |
| $\int_0^{\infty} e^{-3x}\,dx$ | $-\frac{1}{3}(0 - 1)$ | $\frac{1}{3}$ |

In the second row, $e^{2\ln 3} = (e^{\ln 3})^2 = 9$. The last row is an example of a useful fact: $\int_0^\infty e^{-kx}\,dx = \frac{1}{k}$ for $k > 0$.

## Worked examples

**Example 1: a shifted exponent.** $\displaystyle\int e^{3x + 1}\,dx = \frac{1}{3}e^{3x+1} + C$.

**Example 2: a fraction in the exponent.** $\displaystyle\int e^{x/2}\,dx = 2e^{x/2} + C$, since $k = \frac{1}{2}$.

**Example 3: expand first.** $\displaystyle\int (e^x + 1)^2\,dx$. Expand to $e^{2x} + 2e^x + 1$:

$$\int (e^{2x} + 2e^x + 1)\,dx = \frac{1}{2}e^{2x} + 2e^x + x + C$$

**Example 4: simplify a quotient.** $\displaystyle\int \frac{e^{5x}}{e^{3x}}\,dx = \int e^{2x}\,dx = \frac{1}{2}e^{2x} + C$.

**Example 5: growth.** A population grows at a rate $r(t) = 100e^{0.2t}$ per year. The increase over the first 5 years is

$$\int_0^5 100e^{0.2t}\,dt = 500\left(e^{1} - 1\right) \approx 859$$

## Common mistakes

1. **Multiplying instead of dividing.** $\int e^{2x}\,dx$ is not $2e^{2x}$. That is the derivative.
2. **Dividing by $2x$.** The divisor must be a constant. If the exponent is not linear, the shortcut fails.
3. **Applying it to $e^{x^2}$.** $\int e^{x^2}\,dx$ has no elementary antiderivative. See the [integral of e^(x²)](/integrals/e-to-the-x-squared).
4. **Sign slips with negative $k$.** $\int e^{-2x}\,dx = -\frac{1}{2}e^{-2x} + C$. See the [integral of e⁻ˣ](/integrals/e-to-the-negative-x).

## Frequently asked questions

### What is the integral of e^(2x)?

It is $\frac{1}{2}e^{2x} + C$.

### What is the integral of e^(kx)?

It is $\frac{1}{k}e^{kx} + C$ for any nonzero constant $k$. For example, $\int e^{3x}\,dx = \frac{1}{3}e^{3x} + C$ and $\int e^{-4x}\,dx = -\frac{1}{4}e^{-4x} + C$.

### What is the integral of e^(x/2)?

It is $2e^{x/2} + C$. Here $k = \frac{1}{2}$, so $\frac{1}{k} = 2$.

### Is the integral of e^(2x) equal to e^(2x)/(2x)?

No. You divide by the constant $2$, not by $2x$. Check: the derivative of $\frac{1}{2}e^{2x}$ is $e^{2x}$, but the derivative of $\frac{e^{2x}}{2x}$ is not.