Chapter 7 - Section 7.4 - Adding, Subtracting, and Multiplying Radical Expressions - Exercise Set - Page 439: 74

$3x+4\sqrt[]{3x+1}+5$

Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, then, \begin{array}{l} \left( \sqrt[]{3x+1}+2 \right)^2 \\= (\sqrt[]{3x+1})^2+2(\sqrt[]{3x+1})(2)+(2)^2 \\= 3x+1+4\sqrt[]{3x+1}+4 \\= 3x+4\sqrt[]{3x+1}+5 .\end{array}