Use the substitution $u = \ln x$ to find the value of $\int_{\text{e}}^{{{\text{e}}^2}} {\frac{{{\text{d}}x}}{{x\ln x}}}$.

METHOD 1

$\int_{\text{e}}^{{{\text{e}}^2}} {\frac{{{\text{d}}x}}{{x\ln x}}}  = \left[ {\ln (\ln x)} \right]_{\text{e}}^{{{\text{e}}^2}}$     (M1)A1

$= \ln (\ln {{\text{e}}^2}) - \ln (\ln {\text{e}})\;\;\;( = \ln 2 - \ln 1)$     (A1)

$= \ln 2$     A1

[4 marks]

METHOD 2

$u = \ln x,{\text{ }}\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}$     M1

$= \int_1^2 {\frac{{{\text{d}}u}}{u}}$     A1

Note:     Condone absent or incorrect limits here.

$= [\ln u]_1^2$ or equivalent in $x( = \ln 2 - \ln 1)$     (A1)

$= \ln 2$     A1

[4 marks]