We have to evaluate the integral:
[tex]\int4^{3x}dx[/tex]
We start with a substitution:
[tex]\begin{gathered} u=3x \\ \frac{du}{dx}=3\Rightarrow dx=\frac{du}{3} \end{gathered}[/tex]
Then, we can now apply the substitution and then the rule for exponential functions:
[tex]\int4^{3x}dx=\int4^u\frac{du}{3}=\frac{1}{3}\int4^udu[/tex][tex]\frac{1}{3}\int4^udu=\frac{1}{3}\cdot\frac{4^u}{\ln(4)}+C[/tex]
We can replace back with x and write:
[tex]\frac{4^u}{3\ln(4)}=\frac{4^{3x}}{3\ln(4)}[/tex]
Then, the solution to the integral is:
[tex]\int4^{3x}dx=\frac{4^{3x}}{3\ln(4)}+C[/tex]
This result does not match any of the options.
Answer: none of these.
