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JEE Advanced 2026Paper 2
Exam
JEE Advanced
Year
2026
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

JEE Advanced 2026 · Paper 2 · Mathematics · Definite Integration

The value of the definite integral ∫_ 0 ^ 2 1 3^ (x+3)/(2) \, dx is

The value of the definite integral
∫0213x+32 dx\displaystyle\int_{0}^{2} \dfrac{1}{3^{\frac{x+3}{2}}}\, dx
is

Options

  1. A12\dfrac{1}{2}
  2. B13\dfrac{1}{3}
  3. Clog⁡e33\dfrac{\log_e 3}{3}
  4. Dlog⁡e32\dfrac{\log_e 3}{2}

Correct answer

B. 13\dfrac{1}{3}

Step-by-step solution

Let I=∫0213x+3dxI = \int_{0}^{2} \dfrac{1}{3^x + 3} dx

Substitute t=3xt = 3^x, which gives dt=3xln⁡3dx⇒dx=dttln⁡3dt = 3^x \ln 3 dx \Rightarrow dx = \dfrac{dt}{t \ln 3}.

When x=0x = 0, t=1t = 1.

When x=2x = 2, t=9t = 9.

The integral becomes:

I=∫191t+3dttln⁡3=1ln⁡3∫191t(t+3)dtI = \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:

1t(t+3)=13(1t−1t+3)\dfrac{1}{t(t + 3)} = \dfrac{1}{3} \left( \dfrac{1}{t} - \dfrac{1}{t + 3} \right)

I=13ln⁡3∫19(1t−1t+3)dtI = \dfrac{1}{3 \ln 3} \int_{1}^{9} \left( \dfrac{1}{t} - \dfrac{1}{t + 3} \right) dt

I=13ln⁡3[ln⁡t−ln⁡(t+3)]19I = \dfrac{1}{3 \ln 3} [\ln t - \ln(t + 3)]_{1}^{9}

I=13ln⁡3[ln⁡(tt+3)]19I = \dfrac{1}{3 \ln 3} \left[ \ln\left(\dfrac{t}{t + 3}\right) \right]_{1}^{9}

I=13ln⁡3(ln⁡(912)−ln⁡(14))I = \dfrac{1}{3 \ln 3} \left( \ln\left(\dfrac{9}{12}\right) - \ln\left(\dfrac{1}{4}\right) \right)

I=13ln⁡3(ln⁡(34)−ln⁡(14))I = \dfrac{1}{3 \ln 3} \left( \ln\left(\dfrac{3}{4}\right) - \ln\left(\dfrac{1}{4}\right) \right)

I=13ln⁡3ln⁡(3/41/4)=13ln⁡3ln⁡3I = \dfrac{1}{3 \ln 3} \ln\left(\dfrac{3/4}{1/4}\right) = \dfrac{1}{3 \ln 3} \ln 3

I=13I = \dfrac{1}{3}
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