# How do I prove that $\int{e^{xt}dt} = \frac{e^{xt}}{x} + c$?

I'm in differential equations without having taken higher level calculus first

I understand differentiation and integration pretty well but I get surprised calculus-wise periodically.

For example, how is it that $\int{e^{xt}dt} = \frac{e^{xt}}{x} + c$? I must be missing something!

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Try differentiating with respect to t. –  Fredrik Meyer Mar 28 '11 at 22:18
What is the emergency? –  Jonas Meyer Mar 28 '11 at 22:38
You forgot to add +C to the antiderivative to get the most general formula. Especially in a course on differential equations, you need to be aware of constant parameters. –  KCd Mar 28 '11 at 22:44
@Jonas I'm doing laplace transforms now and I couldn't go on if I didn't understand this math. Turned out to be simple. I thought it was going to be complex –  JustcallmeDrago Mar 28 '11 at 23:13
same as $\int e^{3t}dt=e^{3t}/3$ –  yoyo Mar 29 '11 at 0:56

Do you know that $$\int e^u du = e^u +C$$

Now think of integration techniques. Does substitution ring a bell?

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Yeah, now I realize that I was completely missing the fact that I needed x/x to give me the x I needed for the integration, leaving 1/x outside the integral. Thanks –  JustcallmeDrago Mar 28 '11 at 22:24

Let $u=xt$ so that $t=\frac{u}{x}$ and $dt=\frac{du}{x}$. $$\int e^{xt}dt =\int \left(e^u\cdot\frac{du}{x}\right) =\frac{1}{x}\int e^u du =\frac{1}{x}e^u+C =\frac{e^{xt}}{x}+C$$

(Note: the integral you started with is with respect to $t$; as far as the integral is concerned, we're treating $x$ as a constant.)

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Oh! Of course, thanks. Your U substitution made me realize what I was missing in the first place. –  JustcallmeDrago Mar 28 '11 at 22:23

The definition of $e^x$ is:

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

Then the derivative of $e^x$ is

$$\frac{d}{dx} e^x = 0 + 1 + \frac{x}{1!} + \frac{x^2}{2!} + \cdots = e^x$$

by the power rule.

You get your identity by applying the chain rule.

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