Answer
$$A = \frac{1}{4}$$
Work Step by Step
$$\eqalign{
& {\text{The area is given by}} \cr
& A = \int_0^1 {\frac{1}{{{{\left( {3x + 1} \right)}^2}}}} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = \frac{1}{3}\left[ { - \frac{1}{{3x + 1}}} \right]_0^1 \cr
& A = - \frac{1}{3}\left[ {\frac{1}{{3\left( 1 \right) + 1}} - \frac{1}{{3\left( 0 \right) + 1}}} \right] \cr
& {\text{Simplify}} \cr
& A = - \frac{1}{3}\left( {\frac{1}{4} - 1} \right) \cr
& A = \frac{1}{4} \cr} $$