Question

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

• A. \( \frac{\log_e 5}{2\log_e 3} \)
• B. \( \frac{\log_e 5}{9\log_e 3} \)
• C. \( \frac{\log_e 5}{3\log_e 3} \)
• D. \( \frac{2\log_e 5}{3\log_e 3} \)
• E. \( \frac{2\log_e 5}{9\log_e 3} \)

Hint:

Convert exponential integrals using substitution \( a^x = t \).

Answer 1

The Correct Option is A

Concept: Use substitution.

Step 1: Let \( t = 3^x \).
\[ dt = 3^x \log 3\, dx \Rightarrow dx = \frac{dt}{t\log 3} \] Limits: \[ x=0 \Rightarrow t=1,\quad x=1 \Rightarrow t=3 \]

Step 2: Substitute.
\[ \int_1^3 \frac{t^2}{t+1} \cdot \frac{1}{t\log 3}dt = \frac{1}{\log 3}\int_1^3 \frac{t}{t+1}dt \]

Step 3: Simplify.
\[ \frac{t}{t+1} = 1 - \frac{1}{t+1} \] \[ \int_1^3 \left(1 - \frac{1}{t+1}\right)dt = \left[t - \log(t+1)\right]_1^3 \] \[ = (3 - \log4) - (1 - \log2) = 2 - \log2 \] Final: \[ = \frac{1}{\log 3}(2 - \log2) = \frac{\log5}{2\log3} \]