Question

The value of $\underset{e}{\overset{e^e}{\int }}[\frac{1}{\ln x}+\ln (\ln x\left)\right]dx$ is

• A. $1$
• B. $e$
• C. $e^e$
• D. $e^e-e$

Answer 1

The correct option is C $e^e$

$I=\underset{e}{\overset{e^e}{\int }}[\frac{1}{\ln x}+\ln (\ln x\left)\right]dx$
Using integration by parts
$=\int \frac{1}{\ln x}dx+\ln \left(\ln x\right).x-\int \frac{1}{\ln x}\times \frac{1}{x}\times xdx$

$=\int \frac{1}{\ln x}dx+x\ln \left(\ln x\right)-\int \frac{1}{\ln x}dx$
$={\left[x\ln \right(\ln x\left)\right]}_e^{e^e}$
$=e^e$