Proof of exponential function integral 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.
 A: 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.
A: 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.
A: 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
$$\int a^x dx=\int e^{x\ln(a)}dx=\frac1{\ln(a)}e^{x\ln(a)}+k=\frac{a^x}{\ln(a)}+k$$

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

