Basics

Let's start , $\Rightarrow \displaystyle \int \color{#3D99F6}{(3x-4)}^{3}dx$

Let $\color{#3D99F6}{(3x-4)}=\color{#D61F06}{u}$

So ,

$\displaystyle \int \color{#3D99F6}{(3x-4)}^{3}dx \Rightarrow \displaystyle \int \color{#D61F06}{u}^{3}dx$

Now we can re-write it as ,

$\displaystyle \int \color{#D61F06}{u}^{3}dx \Rightarrow \displaystyle \int \color{#BA33D6}{\frac{1}{3}} \color{#D61F06}{u}^{3}dx × \color{#BA33D6}{\frac{du}{dx}}$

$\frac{\color{#BA33D6}{\frac{1}{3}}}{3+1}\color{#D61F06}{u}^{\color{#BA33D6}{3}+1}+C \Rightarrow \frac{1}{12}\color{#D61F06}{u}^{4}+C$

Now replacing $\color{#D61F06}{u}$ with $\color{#3D99F6}{(3x-4)}$ ,

$\frac{1}{12} \color{#3D99F6}{(3x-4)}^{4}+C$

Therefore , $a+b = 12+4 =\huge \boxed{16}$ .