Answer is: option4

Solution:

Interval: \([0, 6]\)

Subdivisions: \(n = 3\)

Subinterval width:

\[ h = \frac{6 - 0}{3} = 2 \]

Subinterval points:
\(x_0 = 0,\ x_1 = 2,\ x_2 = 4,\ x_3 = 6\)

Trapezoidal rule approximation:
\[ \int_a^b f(x)\, dx \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3) \right] \]

Let \(f(x) = \ln(x + 1)\), then:

• \(f(0) = \ln(1)\)
• \(f(2) = \ln(3)\)
• \(f(4) = \ln(5)\)
• \(f(6) = \ln(7)\)

So the approximation becomes: \[ \frac{2}{2} \left[ \ln(1) + 2\ln(3) + 2\ln(5) + \ln(7) \right] \]

Apply log identities:

• \(2\ln(3) = \ln(9)\)
• \(2\ln(5) = \ln(25)\)

So it becomes:
\[ \ln(1) + \ln(9) + \ln(25) + \ln(7) \]

Which matches Option D.