Integrate $\int{2^{2x}} dx$ Integrate $$\int{2^{2x}} dx$$
How do I do it, at first, I thought I treat 2 as $e$ and I will get something like $\dfrac{1}{2} 2^{2x}$, but according to WolframAlpha its supposed to be $\dfrac{4^x}{\lg{4}}$
 A: Well, $$\frac{d}{dx}a^x = a^x \log a.$$ Hence $$\int a^x \, dx = \frac{1}{\log a}a^x+C.$$
A: First, rewrite
$$2^{2x} = (2^2)^x = 4^x.$$
This eliminates the need to do a substitution.  Then, use the rule that
$$\frac{d}{dx} a^x = a^x \ln a,$$
so that
$$\int a^x \,dx = \frac{a^x}{\ln a} + C.$$
If you don't see where this comes from, use logarithmic differentiation.  That is, let $y = a^x$.  Then, $\ln y = \ln a^x = x \ln a$.  Taking the derivative of both sides gives
$$\frac{y'}{y} = \ln a$$
so that
$$y' = y \ln a = a^x \ln a.$$
Now, it is simple to see that
$$\int 2^{2x} \,dx = \int 4^x \,dx = \frac{4^x}{\ln 4} + C.$$
A: $$ I = \int 2^{2x} \mathrm {d}x \tag{1}$$
Let
$$ \begin{align*}
y &=2^{2x} \hspace{5pt} \\
\Rightarrow \ln y &= 2x \ln 2
\end{align*}
$$
Differentiate both sides
$$ 
\begin{align*}
\frac{1}{y} \frac{dy}{dx} &= 2 \ln 2 \\
\frac{dy}{y}&=
2 \ln 2 \hspace{4pt}dx\\ 
&= \ln 2^2 \hspace{4pt} dx
\end{align*}
$$
$$ 
\begin{align*}
dx &= \frac{dy}{y \hspace{4pt} \ln 2^2} 
\end{align*}
$$
Substitute for $x$ and $dx$ in $(1)$
$$ 
\begin{align*}
I &= \int y \frac{dy}{y \hspace{4pt} \ln 2^2}\\
&= \int \frac{1}{\ln 2^2} \mathrm{d}y =  \frac{1}{\ln 2^2} \int \mathrm{d}y =  \frac{1}{\ln 4} \int \mathrm{d}y \\
&= \frac{y}{\ln 4} + C  \hspace{14pt} (\textit{But } y=2^{2x})\\
&= \frac{2^{2x}}{\ln 4} + C 
\end{align*}
$$
