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 $\dfrac{4^x}{\lg{4}}$
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 $\dfrac{4^x}{\lg{4}}$
Well, $$\frac{d}{dx}a^x = a^x \log a.$$ Hence $$\int a^x \, dx = \frac{1}{\log a}a^x+C.$$
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 $$\int 2^{2x} \,dx = \int 4^x \,dx = \frac{4^x}{\ln 4} + C.$$
$$ I = \int 2^{2x} \mathrm {d}x \tag{1}$$ 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 $(1)$
$$ \begin{align*} I &= \int y \frac{dy}{y \hspace{4pt} \ln 2^2}\\ &= \int \frac{1}{\ln 2^2} \mathrm{d}y = \frac{1}{\ln 2^2} \int \mathrm{d}y = \frac{1}{\ln 4} \int \mathrm{d}y \\ &= \frac{y}{\ln 4} + C \hspace{14pt} (\textit{But } y=2^{2x})\\ &= \frac{2^{2x}}{\ln 4} + C \end{align*} $$