Antiderivative of e^x (Plus e^2x, e^-x and x e^x)

\(e^x\) is the friendliest function in calculus: differentiate it and nothing changes, integrate it and nothing changes. The only places students lose points are the variations, like \(e^{2x}\), \(e^{-x}\) and \(x e^x\). This guide covers the main answer, three ways to see why it is true, and a step-by-step ladder of the variations you are most likely to meet.

Three Ways to See Why \(\int e^x\,dx = e^x + C\)

1. Reverse the derivative

The derivative of \(e^x\) is \(e^x\). An antiderivative is any function whose derivative gives you the original, so \(e^x\) works. Adding a constant keeps the derivative the same, so the full answer is \(e^x + C\).

2. Look at the slope

On the graph of \(y = e^x\), the slope at every point equals the height. At \(x = 0\) the height is 1 and the slope is 1. At \(x = 2\) the height is about 7.39 and so is the slope. A function whose slope always matches its height is its own derivative.

3. Use the series

\(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\). Integrate term by term with the power rule and you get \(x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots\), which is \(e^x – 1\). The \(-1\) is absorbed into \(C\).

The Variation Ladder

Each step adds one new idea. Work down the list and you can handle almost any exponential integral.

Step	Integral	Answer	New idea
1	\(\int 5e^x\,dx\)	\(5e^x + C\)	Constants come along
2	\(\int e^{2x}\,dx\)	\(\frac12e^{2x} + C\)	Divide by the inside’s slope
3	\(\int e^{-x}\,dx\)	\(-e^{-x} + C\)	The slope is \(-1\)
4	\(\int e^{3x+1}\,dx\)	\(\frac13e^{3x+1} + C\)	Any straight-line inside
5	\(\int x e^{x^2}\,dx\)	\(\frac12e^{x^2} + C\)	U-substitution
6	\(\int x e^x\,dx\)	\(xe^x – e^x + C\)	Integration by parts
7	\(\int 2^x\,dx\)	\(\frac{2^x}{\ln 2} + C\)	Other bases

Step 2 in detail: \(\int e^{2x}\,dx\)

Let \(u = 2x\), so \(du = 2\,dx\) and \(dx = \frac12\,du\):

\[\int e^{2x}\,dx = \frac12\int e^u\,du = \frac12e^{2x} + C.\]

Check: \(\frac{d}{dx}\left(\frac12e^{2x}\right) = \frac12\cdot2e^{2x} = e^{2x}\). ✓ The \(\frac12\) is there to cancel the 2 that the chain rule creates.

Step 3 in detail: \(\int e^{-x}\,dx\)

Here the inside is \(-x\), with slope \(-1\). Dividing by \(-1\) flips the sign:

\[\int e^{-x}\,dx = -e^{-x} + C.\]

This one shows up constantly in physics and probability, for example in the area under a decay curve: \(\int_0^{\infty} e^{-x}\,dx = 1\).

Step 5 in detail: \(\int x e^{x^2}\,dx\)

The \(x\) outside is half the derivative of \(x^2\). Let \(u = x^2\), so \(x\,dx = \frac12\,du\):

\[\int x e^{x^2}\,dx = \frac12\int e^u\,du = \frac12e^{x^2} + C.\]

Step 6 in detail: \(\int x e^x\,dx\)

Now the \(x\) is not the derivative of the exponent, so substitution fails. Use integration by parts with \(u = x\), \(dv = e^x\,dx\):

\[\int x e^x\,dx = xe^x – \int e^x\,dx = xe^x – e^x + C = e^x(x – 1) + C.\]

Same rule, different letter: \(e^t\), \(e^y\) and \(e^{-t}\)

The letter does not matter, as long as you integrate with respect to that same letter:

• \(\int e^t\,dt = e^t + C\) and \(\int e^y\,dy = e^y + C\)
• \(\int e^{-t}\,dt = -e^{-t} + C\)
• \(\int 2e^x\,dx = 2e^x + C\)

But watch the variable. If you integrate \(e^y\) with respect to \(x\), and \(y\) does not depend on \(x\), then \(e^y\) is a constant: \(\int e^y\,dx = xe^y + C\). See the integral of a constant.

The One You Cannot Do: \(\int e^{x^2}\,dx\)

It looks like step 5, but without the \(x\) in front. There is nothing to cancel the \(2x\) that the chain rule would create, and in fact no formula made of ordinary functions has \(e^{x^2}\) as its derivative. This was proved in the 1800s. You can still find definite values like \(\int_0^1 e^{x^2}\,dx \approx 1.4627\) numerically, for example with the definite integral calculator. More on this in what calculators can and cannot integrate.

The same is true for its relatives. None of these has an elementary antiderivative:

Integral	What you can write instead
\(\int e^{x^2}\,dx\)	\(\frac{\sqrt{\pi}}{2}\operatorname{erfi}(x) + C\) (a special function)
\(\int e^{-x^2}\,dx\)	\(\frac{\sqrt{\pi}}{2}\operatorname{erf}(x) + C\), the famous bell-curve integral
\(\int e^{x^4}\,dx\)	Only a series or a special function

The bell-curve one has a famous definite value: \(\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}\).

Area Under \(e^x\)

\[\int_0^1 e^x\,dx = \Big[e^x\Big]_0^1 = e – 1 \approx 1.718.\]

And over any interval, the area under \(e^x\) from \(a\) to \(b\) is simply \(e^b – e^a\).

Where This Shows Up in Real Life

When something grows or shrinks at a rate proportional to its size, the math leads to \(e^{kt}\). Money with continuous interest, bacteria, cooling coffee and medicine leaving the bloodstream all follow this pattern. Integrating \(e^{kt}\) gives the total amount over time. Try the exponential growth calculator to see these curves with real numbers.

Calculators for This Topic

Practice what you just read. Each tool shows full steps and checks its own answer.

• ∫Antiderivative Calculator – Find any antiderivative with full steps and a check.
• eˣExponential Growth Calculator – Growth and decay, doubling time and a value table.
• ∞Improper Integral Calculator – Check if an integral converges or diverges.

Frequently Asked Questions

Related Guides

• Antiderivatives of all exponential functions, including a^x
• Antiderivative of ln x
• Glossary: exponential growth