Question

The greatest integer less than or equal to $\underset{1}{\overset{2}{\int }}{\log }_2(x^3+1)dx+\underset{1}{\overset{{\log }_{2}9}{\int }}(2^x-1{)}^{\frac{1}{3}}dx$ is

• A. 5
• B. 5.00
• C. 5.0

Answer 1

Answer: (A), (B), (C)

$I=\underset{1}{\overset{2}{\int }}{\log }_2(x^3+1)dx+\underset{1}{\overset{{\log }_{2}9}{\int }}(2^x-1{)}^{\frac{1}{3}}dx$
Let$\underset{1}{\overset{{\log }_{2}9}{\int }}(2^x-1{)}^{\frac{1}{3}}dx=I_2$
Let $2^x-1=t^3$
$\Rightarrow 2^x\ln 2dx=3t^{2}dt$
or $dx=\frac{3t^2}{\ln 2(t^3+1)}dt$
So, $I=\underset{1}{\overset{2}{\int }}{\log }_2(x^3+1)dx+\underset{1}{\overset{2}{\int }}\frac{3t^3}{\ln 2(t^3+1)}dt$
or $I=\underset{1}{\overset{2}{\int }}\left({\log }_2\right(t^3+1)+t.\frac{3t^2}{(t^3+1)\ln 2})dt$
$=t.{\log }_2(t^3+1){|}_1^2$
$=2{\log }_{2}9-1{\log }_{2}2$
$=2{\log }_{2}9-1$
So, $\left[I\right]=5$