Integration (10)

The given integral is

$\large{\displaystyle \int^{e}_{1} \frac{e(1- \ln x)}{x^2 \left(1+\left(\frac{e \ln x}{x}\right)^2 \right)} dx}$

Put $\large{\frac{e \ln x}{x}=t \Rightarrow \frac{e(1- \ln x)}{x^2} dx=dt}$

thus $\large{\displaystyle \int^{1}_{0} \frac{1}{1+t^2} dt \Rightarrow \left( \arctan (t) \right)^{1}_{0}=\frac{\pi}{4}}$

The given integral is a definite integral so on using numeric integration we can solve to an expression using the simpson's rule of numeric integration to get:

(e-1)/6(e+0.1)=0.8 approximately

Now That value is= pi/4 Thus a=4

hence 2016/a=2016/4= 504