Can someone breakdown (ELI$5$) how an integral of a power is evaluated, please? Currently taking Calc $2$ and I find that I get exponential or trig integrals intuitively, except integrals of powers. What am I missing? 
For example:
$$\int\ 3^x{\rm d}x =\frac{3^x}{\ln(3)} + C$$
*Explain like I'm $5$
 A: Basically you need to know two things

  
*
  
*You can express every exponential function of the form $a^x$, with $a>0$, in terms of $e$ by noting that $$e^{x\cdot\ln a}=\left(e^{\ln a}\right)^x=a^x$$
  
*The derivative of a function of the form $f(x)=e^{g(x)}$ is given by
  $$f'(x)=\left(e^{g(x)}\right)'=e^{g(x)}\cdot g'(x)$$
  This follows from the chain rule and the fact that $\left(e^x\right)'=e^x$.
  

Now, combine the first one with the second and therefore notice
$$(a^x)'=(e^{x\cdot\ln a})'=e^{x\cdot\ln a}\cdot(x\cdot\ln a)'=e^{x\cdot\ln a}(\ln a)=a^x\ln a$$
Then, by the Fundamental Theorem of Calculus, it follows immediately

$$\therefore~\int~a^x\ln a~{\rm d}x=a^x+C'\implies\int a^x~{\rm d}x=\frac{a^x}{\ln a}+C$$

Now, set $a=3$ and your result follows.
A: First, what is the derivative of an exponential function?


*

*If $f(x)=e^x$, you know that $\frac{df}{dx}=e^x$.

*If $a>0, a\ne 1$ and $f(x)=a^x=e^{x\ln a}$, then $\frac{df}{dx}=e^{x\ln a}\ln a\text{ (chain rule) }=a^x\ln a$. 

*Now, divide both sides by (constant) $\ln a$: if $f(x)=\frac{a^x}{\ln a}$, then $\frac{df}{dx}=a^x$.

*This means that $\int a^x dx=\frac{a^x}{\ln a}+C$.


Now substitute $a=3$.
A: Try a substitution:
$$3^x=e^{ x\ln 3 }$$
$$\implies u=x\ln 3  \implies \frac {du}{dx}= \ln 3$$
$$\implies dx =\frac {du} {\ln 3}$$
So that
$$I=\int 3^xdx=\int e^{ x\ln 3 }dx$$
$$\implies I=\frac 1 {\ln 3}\int e^udu$$
$$ I=\frac 1 {\ln 3}e^u+C$$
Unsubstitute u $(=x \ln 3)$:
$$ I=\frac {3^x} {\ln 3}+C$$
