How can I compute $\int\frac1{e^{ct}} dt $

Question

$$ \displaystyle \int \dfrac{1}{e^{ct}} dt $$

How do I do this, I've tried substitution. Rewrite the above integral as $\displaystyle \int e^{-ct} dt $, then we have $u=-ct$, but at this point derivatives and integrals are confusing me. If we want $[e^u]'$, we just have $[e^u]u'$, but for integrals I've forgotten how to tackle this.

Answer 1

Let $u=-ct$ then $du = -c\,dt$ or alternatively: $\,\,-\frac{1}{c}du = dt\,\,$ thus your integral becomes

$$\int e^{-ct} dt = -\int e^{u}\frac{1}{c}du = -\frac{1}{c}\int e^{u}du.$$

Can you take it from here?

Answer 2

No need for the u-sub. The anti derivative is the e-power again. You just need to "account for" the derivative of your exponent, which is $-c$. So divide by that and you are done

Answer 3

Please change "numerator" to "denominator" in the title of your post, because the way you have written the integral the $e$ is in the denominator. In a fraction like $\frac35$ the $3$ is the numerator and the $5$ is the denominator; the numerator tells us how many of them there are (there are three of them), the denominator tells us what they are (they are fifths).

Now about that confusing integral $\int e^{-ct}dt$. I want a function whose derivative is $e^{-ct}$. I know how to differentiate exponential functions, but integration confuses me. What I do is use the "guess and correct" method. I know that the integral of $e^t$ is $e^t$; $e^{-ct}$ is sort of like $e^t$; my first guess is$$\int e^{-ct}dt=e^{-ct}.$$I check by differentiating:$$\frac d{dt}e^{-ct}=-ce^{-ct}.$$*Oops*, I got an unwanted factor of $-c$. I correct for that by dividing my first guess by $-c$:$$\frac d{dt}\frac1{-c}e^{-ct}=\frac1{-c}\cdot-ce^{-ct}=e^{-ct}.$$That worked!! So the correct answer is$$\int e^{-ct}dt=\frac{e^{-ct}}{-c}.$$