Can someone breakdown (ELI$5$) how an integral of a power is evaluated, please?

Question

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 $5$

Answer 1

Basically you need to know two things

1. You can express every exponential function of the form $a^x$, with $a>0$, in terms of $e$ by noting that $$e^{x\cdot\ln a}=\left(e^{\ln a}\right)^x=a^x$$
2. 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.

Answer 2

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 $a=3$.

Answer 3

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 $(=x \ln 3)$: $$ I=\frac {3^x} {\ln 3}+C$$