JEE maths#9

$\large \displaystyle \int_e^{e^{2010}} \frac1x \left [ 1+ \frac{1 - x}{\ln(x) \ln \left ( \frac{x}{\ln(x)} \right ) } \right ] dx$

$\color{#20A900} \large \displaystyle \ln(x) = u \rightarrow \frac1x dx = du$

$\large \displaystyle = \int_1^{2010} \left ( 1 + \frac{1-u}{u(u - \ln(u))} \right ) du$

$\large \displaystyle = 2009 + \int_1^{2010} \left (\frac{1-u}{u(u - \ln(u))} \right ) du$

$\large \displaystyle = 2009 + \int_1^{2010} \left (\frac{\frac1u -1}{u - \ln(u)} \right ) du$

$\color{#20A900} \large \displaystyle u - \ln(u) = v \rightarrow \left ( 1 - \frac1u \right ) du = dv$

$\large \displaystyle = 2009 - \int_1^{2010-\ln(2010)} \frac{dv}{v}$

$\large \displaystyle = 2009 - \ln(2010 - \ln(2010))$

$\color{#3D99F6} \large \displaystyle a = 2010 , b = 2009, \boxed{\large \displaystyle a-b = 1}$

$I= \displaystyle\int_{e}^{e^{2010}} \dfrac{1}{x} \left(1+\frac{1-\ln x}{\ln x\ln(\frac{x}{\ln x})} \right)dx\\ = \ln x\Big|_e^{e^{2010}}+\displaystyle\int_{e}^{e^{2010}} \frac{1-\ln x}{\ln x\ln(\frac{x}{\ln x})}\cdot\frac{dx}{x} \\ =2009+\displaystyle\int_{e}^{e^{2010}} \frac{1-\ln x}{\ln x\ln(\frac{x}{\ln x})}\cdot\frac{dx}{x}$

Now, take $\cfrac{x}{\ln x}=y \implies \cfrac{\ln x-1}{(\ln x)^2}dx=dy$

$\implies \displaystyle\int_{e}^{e^{2010}} \frac{1-\ln x}{\ln x\ln(\frac{x}{\ln x})}\cdot\frac{dx}{x}= -\displaystyle\int_{e}^{\big(\frac{e^{2010}}{2010}\big)} \frac{dy}{y\ln y}=-\displaystyle\int_{e}^{\big(\frac{e^{2010}}{2010}\big)} \frac{d(\ln y)}{\ln y}=-\ln(\ln y) \Big|_{e}^{\big(\frac{e^{2010}}{2010}\big)}=-\ln(2010-\ln2010)$

So, $I=2009-\ln(2010-\ln2010)\implies \boxed{a-b=1}$