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I fully apologize if this is a duplicate, but I really can’t find the answer to this online here. But a proof of the rule

$$\int a^x dx= \frac{a^x}{\ln(a)}+C$$

would be really helpful. I need it for something I’m doing. This is all I need. I have no sources on the problem as I want the proof myself, but I use this rule so much that it struck me as second nature when I wanted to solve it! Any possible answers would be appreciated.

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    $\begingroup$ It's quite simple. $(a^x)' = a^x\cdot \ln(a)$, and the rule follows. $\endgroup$ – Jakobian Dec 16 '18 at 23:21
  • $\begingroup$ @Jakobian : The only real answer. +1. $\endgroup$ – MPW Dec 16 '18 at 23:46
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I assume it's kosher to use the exponential integral with base $e$, i.e. $\int e^x dx = e^x +C$? Or, more generally, for a constant $k$,

$$\int e^{kx}dx = \frac{1}{k}e^{kx}+C$$

If so, then note:

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

This is mostly just manipulation of various logarithm properties: namely,

$$e^{\ln(x)} = x$$ $$\ln(a^b) = b \ln(a)$$

Also, a nitpick: the integral in your question, OP, needs a $+C$ after it, since indefinite integration introduces an arbitrary constant.

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  • $\begingroup$ OK, so I know this already seems taken for granted, I guess (because it was used in all the other proofs I saw online), can you please furthermore prove ∫e^kx dx=1/k(e^kx)+C? $\endgroup$ – Math Bob Dec 16 '18 at 23:51
  • $\begingroup$ Personally I like proving it by using the power series definition for $e^{kx}$, i.e. $$e^{kx} = \sum_{n=0}^\infty \frac{(kx)^n}{n!}$$ but that involves a few technicalities regarding how infinite summations work under differentiation/integration. You could always just go "well, see, $$\frac{d}{dx} \frac{e^{kx}}{k} + C= k \frac{e^{kx}}{k} +0= e^{kx}$$ so $e^{kx}/k + C$ must be the antiderivative." $\endgroup$ – Eevee Trainer Dec 17 '18 at 0:18
  • $\begingroup$ Okay. I have taken this in so thanks for the answer! $\endgroup$ – Math Bob Dec 17 '18 at 4:04
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Well this one can be found within every good integration table $($e.g. take at look at this$)$. Anyway it is actually not that hard to compute therefore I will demonstrate it for you. First of all the basic property of any exponential function with a basis $a$ is that it can be rewritten in terms of $e$ in the following way

$$a^x=\left(e^{\ln(a)}\right)^x=e^{x\ln(a)}$$

Now we know that $e^x$ remains $e^x$ after integration aswell as after differentiation. Adding a constant $c$ before the $x$ within the exponent yields to

$$\frac d{dx}e^{cx}=ce^{cx}\text{ and }\int e^{cx}dx=\frac1ce^{cx}+k$$

From hereon we are basically done since $\ln(a)$ can be seens as a constant while integrating. So plugging this together leads to

$$\int a^x dx=\int e^{x\ln(a)}dx=\frac1{\ln(a)}e^{x\ln(a)}+k=\frac{a^x}{\ln(a)}+k$$

$$\int a^xdx=\frac{a^x}{\ln(a)}+k$$

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Just differentiate the right hand side and see what you get. Note that $$ \frac{d}{dx}\left(\frac{a^x}{\ln a}\right)=\frac{1}{\ln a}\frac{d}{dx}(e^{x\ln a})=\frac{1}{\ln a}\times e^{x\ln a}\times \ln a=a^x $$ where we used the chain rule in the second equality.

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