The value of the definite integral $\int_{0}^{2}\frac{1}{3^{x}+3}\,\mathrm{d}x$ is

The value of the definite integral ∫₀²1/(3ˣ+3) dx is $\displaystyle\int_0^2 \dfrac{1}{3^{x}+3}\,\mathrm{d}x$ $\text{To find: the value of the integral}$ $(1)\ \dfrac{1}{2} \quad (2)\ \dfrac{1}{3} \quad (3)\ \dfrac{\log_e 3}{3} \quad (4)\ \dfrac{\log_e 3}{2}$ $\int_0^{a} f(x)\,dx=\int_0^{a} f(a-x)\,dx,\quad a=2$ $f(x)=\dfrac{1}{3^{x}+3}$ $f(2-x)=\dfrac{1}{3^{2-x}+3}=\dfrac{1}{\dfrac{9}{3^{x}}+3}$ $f(2-x)=\dfrac{3^{x}}{9+3\cdot 3^{x}}=\dfrac{3^{x}}{3(3^{x}+3)}$ $f(x)+f(2-x)=\dfrac{1}{3^{x}+3}+\dfrac{3^{x}}{3(3^{x}+3)}=\dfrac{3+3^{x}}{3(3^{x}+3)}=\dfrac{1}{3}$ $2I=\int_0^2\big(f(x)+f(2-x)\big)\,dx=\int_0^2\dfrac{1}{3}\,dx$ $2I=\dfrac{1}{3}\times 2=\dfrac{2}{3} \Rightarrow I=\dfrac{1}{3}$ $\int_0^2\dfrac{dx}{3^{x}+3}=\left[\dfrac{1}{3\ln 3}\ln\dfrac{3^{x}}{3^{x}+3}\right]_0^2$ $=\dfrac{1}{3\ln 3}\left(\ln\dfrac{3}{4}-\ln\dfrac{1}{4}\right)=\dfrac{\ln 3}{3\ln 3}=\dfrac{1}{3}$ $\dfrac{1}{3}\ \Rightarrow\ \text{option }(2)$ $\boxed{\dfrac{1}{3}}$