The value of the definite integral ${\int }_0^2\frac{1}{3^x+3}dx$ is

The correct option is $\frac{1}{3}$.

We are required to evaluate the definite integral:

$I={\int }_0^2\frac{1}{3^x+3}dx$

Step 1: Simplify the integrand using properties of definite integrals

We use the definite integral property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$. Here, $a=0$ and $b=2$, so we substitute $x$ with $2-x$:

$I={\int }_0^2\frac{1}{3^{2-x}+3}dx$

Rewriting $3^{2-x}$ as $\frac{3^2}{3^x}=\frac{9}{3^x}$:

$I={\int }_0^2\frac{1}{\frac{9}{3^x}+3}dx$

$I={\int }_0^2\frac{3^x}{9+3·3^x}dx$

Factoring out 3 from the denominator:

$I={\int }_0^2\frac{3^x}{3(3+3^x)}dx=\frac{1}{3}{\int }_0^2\frac{3^x}{3^x+3}dx$

Thus, we have:

$3I={\int }_0^2\frac{3^x}{3^x+3}dx$

Step 2: Add the two representations of the integral

Now, add $I$ and $3I$:

$I+3I={\int }_0^2\frac{1}{3^x+3}dx+{\int }_0^2\frac{3^x}{3^x+3}dx$

$4I={\int }_0^2\frac{1+3^x}{3^x+3}dx$

Step 3: Evaluate the combined integral

Notice that $1+3^x=3^x+1$ and $3^x+3=(3^x+1)+2$. Alternatively, substitute directly or split into simpler parts:

${\int }_0^2\frac{3^x+1}{3^x+3}dx={\int }_0^2(1-\frac{2}{3^x+3})dx$

$4I={\int }_0^{2}1dx-2{\int }_0^2\frac{1}{3^x+3}dx$

Since ${\int }_0^2\frac{1}{3^x+3}dx=I$, we have:

$4I={\left[x\right]}_0^2-2I$

$4I=(2-0)-2I$

$4I+2I=2$

$6I=2$

$I=\frac{2}{6}=\frac{1}{3}$

Therefore, the value of the given integral is $\frac{1}{3}$.