JEE Advanced 2026 Paper 2Question 1
The value of the definite integral $$\int_0^2 \frac{1}{3^x+3}\,dx$$ is
Definite Integration
Solution
Concept. Write $\frac{1}{3^x+3}=\frac13\Big(1-\frac{3^x}{3^x+3}\Big)$; the second part is $\frac{g'}{g}$. Formulas used. (1) $\int\dfrac{3^x}{3^x+3}\,dx=\dfrac{\ln(3^x+3)}{\ln3}$Step 1 - Split. $$\int_0^2\frac{dx}{3^x+3}=\frac13\int_0^2\Big(1-\frac{3^x}{3^x+3}\Big)dx.$$ Step 2 - Use (1). $$=\frac13\Big[x-\frac{\ln(3^x+3)}{\ln3}\Big]_0^2=\frac13\Big(2-\frac{\ln12-\ln4}{\ln3}\Big)=\frac13(2-1)=\frac13.$$ Answer: (B)