The value of the definite integral \({\int }_{0}^{2}\frac{1}{{3}^{x}+3}dx\) is [JEE Advanced 2026]

Given

\(\displaystyle I=\int_0^2 \frac{1}{3^x+3}\,dx\)

Let \(3^x=t\)

\(\Rightarrow dx=\dfrac{dt}{t\ln3}\)

When \(x=0,\; t=1\)

When \(x=2,\; t=9\)

Therefore,

\(\displaystyle I=\frac1{\ln3}\int_1^9 \frac{1}{t(t+3)}\,dt\)

\(=\frac1{\ln3}\int_1^9\left(\frac1{3t}-\frac1{3(t+3)}\right)dt\)

\(=\frac1{3\ln3}\left[\ln t-\ln(t+3)\right]_1^9\)

\(=\frac1{3\ln3}\left[\ln\frac{9}{12}-\ln\frac14\right]\)

\(=\frac1{3\ln3}\ln\left(\frac34\cdot4\right)\)

\(=\frac1{3\ln3}\ln3\)

\(=\frac13\)

Hence,

\(\displaystyle \int_0^2 \frac{1}{3^x+3}\,dx=\frac13\)