JEE Advanced 2026 Paper 2, Mathematics Section 1 Q4: Methods of Integration

Rewrite the integrand by multiplying top and bottom by $3^{-x}$: $\frac{1}{3^x+3}=\frac{3^{-x}}{1+3\cdot 3^{-x}}$.

Substitute $t=3^{-x}$, so $dt=-3^{-x}{\log }_{e}3dx$, that is $3^{-x}dx=-\frac{dt}{{\log }_{e}3}$.

Limits: $x=0\Rightarrow t=1$; $x=2\Rightarrow t=\frac{1}{9}$.

$I={\int }_1^{1/9}\frac{-1}{{\log }_{e}3}\cdot \frac{dt}{1+3t}=\frac{1}{{\log }_{e}3}{\int }_{1/9}^1\frac{dt}{1+3t}$.

$=\frac{1}{3{\log }_{e}3}[{\log }_e(1+3t){]}_{1/9}^1=\frac{1}{3{\log }_{e}3}\left({\log }_{e}4-{\log }_e\frac{4}{3}\right)=\frac{{\log }_{e}3}{3{\log }_{e}3}=\frac{1}{3}$.