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(D) Let $f(x) = \log_2(x^3+1)$. Then $y = \log_2(x^3+1) \implies x^3+1 = 2^y \implies x = (2^y-1)^{1/3}$.Thus,$f^{-1}(x) = (2^x-1)^{1/3}$.The given in

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$5$

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(D) Let $f(x) = \log_2(x^3+1)$. Then $y = \log_2(x^3+1) \implies x^3+1 = 2^y \implies x = (2^y-1)^{1/3}$.Thus,$f^{-1}(x) = (2^x-1)^{1/3}$.The given integral is of the form $\int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = b f(b) - a f(a)$.Here $a=1$,$b=2$. $f(1) = \log_2(1^3+1) = \log_2 2 = 1$. $f(2) = \log_2(2^3+1) = \log_2 9$.So,the integral value is $2 \cdot f(2) - 1 \cdot f(1) = 2 \log_2 9 - 1$.Since $8 Multiplying by $2$,we get $6 Subtracting $1$,we get $5 Specifically,$2 \log_2 9 - 1 = \log_2 81 - 1 = \log_2 81 - \log_2 2 = \log_2(40.5)$.Since $2^5 = 32$ and $2^6 = 64$,$5 The greatest integer less than or equal to this value is $5$.

The greatest integer less than or equal to $\int_1^2 \log _2(x^3+1) dx + \int_1^{\log_2 9} (2^x-1)^{1/3} dx$ is . . . . .

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