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

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    If you just divide your answer by $\ln(2)$ then you'll get Wolfram's answer. And you're right, $2$ and $e$ are both constants and they almost share the same properties, at least relative to integration of the corresponding exponentials. $e^x$ is a bit cleaner though since dividing by $\ln(e)$ amounts to dividing by $1$..2012-05-08

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Well, $\frac{d}{dx}a^x = a^x \log a.$ Hence $\int a^x \, dx = \frac{1}{\log a}a^x+C.$

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    These things are always a bit "circular". You need to know that $\lim_{x \to 0} \frac{a^x-1}{x}=\log a,$ otherwise you are out of luck. But this is the definition of the derivative of $x \mapsto a^x$ at $x=0$. Exponential and logarithms should be either defined correctly, or taken as granted to some extent.2012-05-08
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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 $\int 2^{2x} \,dx = \int 4^x \,dx = \frac{4^x}{\ln 4} + C.$

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$ 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*} $