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Definite Integration

Section: Chemistry

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Question : 1 of 47

Marks: +1, -0

The value of the definite integral

\int\limits_{0}^{2} \frac{1}{3^x + 3} dx

[JEE Adv 2026 P2]

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{\log_e 3}{3}$
$\frac{\log_e 3}{2}$

Solution:

Concept:

Use substitution

3^x = t

to convert the integral into a rational function, then apply partial fractions and integrate.

Explanation:

Let

I = \int_{0}^{2} \frac{1}{3^x + 3} \, dx

.

Set

t = 3^x \Rightarrow dt = 3^x \ln 3 \, dx = t \ln 3 \, dx \Rightarrow dx = \frac{dt}{t \ln 3}

.

When

x = 0

,

t = 1

; when

x = 2

,

t = 9

.

Thus

I = \int_{1}^{9} \frac{1}{t+3} \cdot \frac{1}{t \ln 3} \, dt = \frac{1}{\ln 3} \int_{1}^{9} \frac{1}{t(t+3)} \, dt

.

Use partial fractions:

\frac{1}{t(t+3)} = \frac{A}{t} + \frac{B}{t+3}

.

Multiply:

1 = A(t+3) + Bt = (A+B)t + 3A

.

Comparing coefficients:

A+B = 0

,

3A = 1 \Rightarrow A = \frac{1}{3}

,

B = -\frac{1}{3}

.

Hence

\frac{1}{t(t+3)} = \frac{1}{3}\left(\frac{1}{t} - \frac{1}{t+3}\right)

.

Then

I = \frac{1}{3\ln 3} \int_{1}^{9} \left(\frac{1}{t} - \frac{1}{t+3}\right) dt = \frac{1}{3\ln 3} \left[ \ln t - \ln(t+3) \right]_{1}^{9}

.

Simplify:

I = \frac{1}{3\ln 3} \left[ \ln \frac{t}{t+3} \right]_{1}^{9} = \frac{1}{3\ln 3} \left( \ln \frac{9}{12} - \ln \frac{1}{4} \right) = \frac{1}{3\ln 3} \left( \ln \frac{3}{4} - \ln \frac{1}{4} \right)

.

Since

\ln a - \ln b = \ln \frac{a}{b}

, we get

I = \frac{1}{3\ln 3} \ln 3 = \frac{1}{3}

.

Answer:

\frac{1}{3}

(Option B).

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