Shown below is the curve defined by the equation y = log(x + 1) for x…

The correct answer is Option (1) → $6 \log(2) - 2$

We have, $y = \log(x + 1)$

$\text{Required Area} = \int_{1}^{3} \log(x + 1) dx \text{}$

$= \int_{1}^{3} \log(x + 1) \cdot 1 \, dx \text{}$

$= [\log(x + 1) \cdot x]_{1}^{3} - \int_{1}^{3} x \cdot \frac{1}{(x + 1)} dx \text{}$

$= (3 \log 4 - \log 2) - \int_{1}^{3} \left( 1 - \frac{1}{x + 1} \right) dx \text{}$

$= (6 \log 2 - \log 2) - [x]_{1}^{3} + [\log(x + 1)]_{1}^{3} \text{}$

$= 5 \log 2 - 2 + (\log 4 - \log 2) \text{}$

$= 5 \log 2 - 2 + (2 \log 2 - \log 2) \text{}$

$= 6 \log 2 - 2 \text{}$