Question

If $I={\int }_{\alpha }^{\beta }[\log \log x+\frac{1}{(\log x{)}^2}]dx,$ then $I$
equals

• A. $\alpha \log \log \alpha -\beta \log \log \beta$
• B. $\frac{1}{\alpha }-\frac{1}{\beta }+\log \log \alpha -\log \log \beta$
• C. $\frac{\beta -\alpha }{\alpha \beta }+\alpha \log \log \alpha -\beta \log \log \beta$
• D. none of these

Answer 1

The correct option is D none of these

Put $\log x=t,$ or $x=e^t,$ so that
$I={\int }_a^b[\log t+\frac{1}{t^2}]e^{t}dt$
where $a=\log \alpha ,b=\log \beta$
$={\int }_a^b(\log t+\frac{1}{t}+(-\frac{1}{t})+\frac{1}{t^2})e^{t}dt$
$=(\log t-\frac{1}{t})e^t{]}_a^b$
$[use\int e^x(f\left(x\right)+f^{\prime }\left(x\right))=e^{x}f(x\left)\right]$
$=(\log b-\frac{1}{b})e^b-(\log a-\frac{1}{a})e^a$
$=(\log \log \beta -\frac{1}{\log \beta })\beta -(\log \log \alpha -\frac{1}{\log \alpha })\alpha$