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Exponential and log integrals

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.

Exponential and log integrals Updated 2 min read

At a glance

\int e^{2x}\,dx = \frac{1}{2}e^{2x} + C
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
On this page
  1. Graph of e²ˣ and its antiderivative
  2. How to integrate e²ˣ step by step
  3. Check by differentiating
  4. Definite integrals
  5. Worked examples
  6. Common mistakes

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

y = e^(2x) and its antiderivative ½e^(2x)
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.

How to integrate e²ˣ step by step

Use u-substitution.

  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.
  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 brings out a factor k, and the \frac{1}{k} cancels it. See the derivative of e²ˣ 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).
  4. Sign slips with negative k. \int e^{-2x}\,dx = -\frac{1}{2}e^{-2x} + C. See the integral of e⁻ˣ.

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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.

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