Question

The greatest integer less than or equal to $\int\limits_1^2 \log _2\left(x^3+1\right) d x+\int\limits_1^{\log _2 9}\left(2^x-1\right)^{\frac{1}{3}} dx$ is _____.

Answer 1

The correct answer is 5.

Answer 2

Let's evaluate the given integral step-by-step, using the provided steps and explanations.

Given:
\[ I = 2 \int_{1}^{2} \log_2(x^3 + 1) \, dx + \log_2 9 \int_{1}^{3} (2x - 1) \, dx \]

First, we'll address the substitution for the second integral.

 Evaluating \(\log_2 9 \int_{1}^{3} (2x - 1) \, dx\):

Let \( 2x - 1 = t^3 \).

Then, \( dx = \frac{3t^2}{2 \ln 2 (t^3 + 1)} dt \).

Substituting this into the integral:

\[ \int_{1}^{3} \log_2(2x - 1) \, dx = \int_{1}^{3} \log_2(t^3 + 1) \frac{3t^2}{2 \ln 2 (t^3 + 1)} dt \]

This becomes:

\[ \int_{1}^{3} \log_2(t^3 + 1) \frac{3t^2}{2 \ln 2 (t^3 + 1)} dt \]
\[ = \int_{1}^{3} \left( \frac{\log_2(t^3 + 1)}{2 \ln 2} + \frac{t \cdot 3t^2}{2 \ln 2 (t^3 + 1)} \right) dt \]

Now we evaluate:

\[ \int_{1}^{3} \left( \frac{\log_2(t^3 + 1)}{2 \ln 2} + \frac{3t^2}{2 \ln 2 (t^3 + 1)} \right) dt \]

The result is simplified to:

\[ \left[ t \log_2(t^3 + 1) \right]_1^3 \]

Evaluate this:

\[ 3 \log_2(27 + 1) - 1 \log_2(2) \]
\[ = 3 \log_2(28) - \log_2(2) \]
\[ = 3 \log_2(28) - 1 \]

We know \(\log_2(28) = 2 + \log_2(7)\). So,

\[ 3(2 + \log_2(7)) - 1 \]
\[ = 6 + 3 \log_2(7) - 1 \]
\[ = 5 + 3 \log_2(7) \]

Therefore, the simplified integral I is,

\[ I = 2 \int_{1}^{2} \log_2(x^3 + 1) \, dx + \log_2 9 \int_{1}^{3} (2x - 1) \, dx \]

Evaluating the integral:

\[ \int_{1}^{2} \log_2(x^3 + 1) \, dx + \int_{1}^{3} \log_2(2x - 1) \, dx \]
\[ = 2 \log_2 9 - 1 \]
\[ = 2 \log_2 9 - 1 \]

Thus, \([I] = 5\).

 Conclusion:
Therefore, the greatest integer less than or equal to \( I \) is indeed:
\[ \boxed{5} \]