Integrate $\int {{3^x}dx} $ .

Hint: This question deals with the natural logarithm and the application of it’s properties. To solve this integral, it should be converted into a simpler form so that it can be integrated easily and for that the basic knowledge of natural logarithm and it’s properties is required. The natural log can be represented in more than one way, like: ${\log _e}x = {\ln _e}x = \ln x$ , they all mean the same. Some basic properties of natural logarithm are: $\left( 1 \right)\ln 1 = 0$ $\left( 2 \right)\ln e = 1$ $\left( 3 \right)\ln {e^x} = x$ $\left( 4 \right)\ln y = \ln x$ , means $x = y$ .

Complete step by step answer:
This question is a bit different since here $x$ is in the power of $3$ . Therefore we will try to replace $3$ by something else in such a way that it can be easily integrated.
Let us understand one concept for that, where we will prove that; ${e^{\ln x}} = x$ .
To prove this, let $y = {e^{\ln x}}$ $......\left( 1 \right)$
By taking log on both sides, we get;
$ \Rightarrow \ln \left( y \right) = \ln {e^{\ln \left( x \right)}}$
By the standard logarithmic rule, we know that; ${\log _b}{a^n} = n{\log _b}a$
Applying it in the above equation, we get;
$ \Rightarrow \ln \left( y \right) = \ln \left( x \right)\ln \left( e \right)$
$\because \ln \left( e \right) = 1$ (Because exponential and logarithmic functions are inverse of each other)
$ \Rightarrow \ln \left( y \right) = \ln \left( x \right)$
Here, notice that the logarithm of the two variables are equal, therefore the variables will also be equal;
$ \Rightarrow y = x$
Replace the x value by $y$ in equation $\left( 1 \right)$, we get;
$ \Rightarrow x = {e^{\ln x}}$
Hence proved that ${e^{\ln x}} = x$.
Now, we will try to use this identity to solve our given question;
$ = \int {{3^x}dx} $ $......\left( 2 \right)$
We can write $3$ as; ${e^{\ln 3}} = 3$ using identity ${e^{\ln x}} = x$ ;
Putting the value of $3$ in equation $\left( 2 \right)$, we get;
$ \Rightarrow \int {{{\left( {{e^{\ln 3}}} \right)}^x}} dx$
On further simplification;
$ \Rightarrow \int {{e^{\ln 3x}}} dx$
By identity $\int {{e^x}dx} = {e^x}$
Here $x$ is $\ln 3x$, so our integral reduces to;
$ \Rightarrow \dfrac{1}{{\ln 3}}{\left( {{e^{\ln 3}}} \right)^x} + C$ ( where $C$ is the integration constant )
And we assumed that ${e^{\ln 3}} = 3$ ;
$ \Rightarrow \dfrac{{{3^x}}}{{\ln 3}} + C$
Therefore, the correct answer for this question is $\dfrac{{{3^x}}}{{\ln 3}} + C$

Note: This question tests our knowledge about the understanding of logarithmic functions and basic identities. Some of the most frequently used and important logarithmic identities are: $\left( 1 \right)$ Logarithmic additive identity is given by $ \Rightarrow {\log _b}\left( {xy} \right) = {\log _b}x + {\log _b}y$. $\left( 2 \right)$ Logarithmic subtractive identity : ${\log _b}\left( {\dfrac{x}{y}} \right) = {\log _b}x - {\log _b}y$. $\left( 3 \right)$ ${\log _b}{a^n} = n{\log _b}a$ .