JEE Advanced 2026 · Paper 2, Mathematics question 4

Write $\dfrac1{3^x+3}=\dfrac13\cdot\dfrac1{3^{x-1}+1}$ and put $u=x-1$:
$$I=\frac13\int_{-1}^{1}\frac{du}{3^u+1}.$$
With $h(u)=\frac1{3^u+1}$, $h(u)+h(-u)=\frac1{3^u+1}+\frac{3^u}{1+3^u}=1$, so the integral over $[-1,1]$ is half the length, $1$.

Hence $I=\dfrac13$.