Q22 / 54 · JEE Advanced 2026 (Paper 2) · Mathematics
The value of the definite integral
$\displaystyle\int_{0}^{2} \dfrac{1}{3^{\frac{x+3}{2}}}\, dx$
is
$\displaystyle\int_{0}^{2} \dfrac{1}{3^{\frac{x+3}{2}}}\, dx$
is
- A$\dfrac{1}{2}$
- Correct answerB$\dfrac{1}{3}$
- C$\dfrac{\log_e 3}{3}$
- D$\dfrac{\log_e 3}{2}$
Solution
Let $I = \int_{0}^{2} \dfrac{1}{3^x + 3} dx$
Substitute $t = 3^x$, which gives $dt = 3^x \ln 3 dx \Rightarrow dx = \dfrac{dt}{t \ln 3}$.
When $x = 0$, $t = 1$.
When $x = 2$, $t = 9$.
The integral becomes:
$I = \int_{1}^{9} \dfrac{1}{t + 3} \dfrac{dt}{t \ln 3} = \dfrac{1}{\ln 3} \int_{1}^{9} \dfrac{1}{t(t + 3)} dt$
Using partial fractions:
$\dfrac{1}{t(t + 3)} = \dfrac{1}{3} \left( \dfrac{1}{t} - \dfrac{1}{t + 3} \right)$
$I = \dfrac{1}{3 \ln 3} \int_{1}^{9} \left( \dfrac{1}{t} - \dfrac{1}{t + 3} \right) dt$
$I = \dfrac{1}{3 \ln 3} [\ln t - \ln(t + 3)]_{1}^{9}$
$I = \dfrac{1}{3 \ln 3} \left[ \ln\left(\dfrac{t}{t + 3}\right) \right]_{1}^{9}$
$I = \dfrac{1}{3 \ln 3} \left( \ln\left(\dfrac{9}{12}\right) - \ln\left(\dfrac{1}{4}\right) \right)$
$I = \dfrac{1}{3 \ln 3} \left( \ln\left(\dfrac{3}{4}\right) - \ln\left(\dfrac{1}{4}\right) \right)$
$I = \dfrac{1}{3 \ln 3} \ln\left(\dfrac{3/4}{1/4}\right) = \dfrac{1}{3 \ln 3} \ln 3$
$I = \dfrac{1}{3}$
Substitute $t = 3^x$, which gives $dt = 3^x \ln 3 dx \Rightarrow dx = \dfrac{dt}{t \ln 3}$.
When $x = 0$, $t = 1$.
When $x = 2$, $t = 9$.
The integral becomes:
$I = \int_{1}^{9} \dfrac{1}{t + 3} \dfrac{dt}{t \ln 3} = \dfrac{1}{\ln 3} \int_{1}^{9} \dfrac{1}{t(t + 3)} dt$
Using partial fractions:
$\dfrac{1}{t(t + 3)} = \dfrac{1}{3} \left( \dfrac{1}{t} - \dfrac{1}{t + 3} \right)$
$I = \dfrac{1}{3 \ln 3} \int_{1}^{9} \left( \dfrac{1}{t} - \dfrac{1}{t + 3} \right) dt$
$I = \dfrac{1}{3 \ln 3} [\ln t - \ln(t + 3)]_{1}^{9}$
$I = \dfrac{1}{3 \ln 3} \left[ \ln\left(\dfrac{t}{t + 3}\right) \right]_{1}^{9}$
$I = \dfrac{1}{3 \ln 3} \left( \ln\left(\dfrac{9}{12}\right) - \ln\left(\dfrac{1}{4}\right) \right)$
$I = \dfrac{1}{3 \ln 3} \left( \ln\left(\dfrac{3}{4}\right) - \ln\left(\dfrac{1}{4}\right) \right)$
$I = \dfrac{1}{3 \ln 3} \ln\left(\dfrac{3/4}{1/4}\right) = \dfrac{1}{3 \ln 3} \ln 3$
$I = \dfrac{1}{3}$
Hence, the correct answer is option: 'B'.