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 $\dfrac{4^x}{\lg{4}}$

$$ I = \int 2^{2x} \mathrm {d}x \tag{1}$$
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 $(1)$
$$ 
\begin{align*}
I &= \int y \frac{dy}{y \hspace{4pt} \ln 2^2}\\
&= \int \frac{1}{\ln 2^2} \mathrm{d}y =  \frac{1}{\ln 2^2} \int \mathrm{d}y =  \frac{1}{\ln 4} \int \mathrm{d}y \\
&= \frac{y}{\ln 4} + C  \hspace{14pt} (\textit{But } y=2^{2x})\\
&= \frac{2^{2x}}{\ln 4} + C 
\end{align*}
$$
