Logarithmic Integration!

Put $lnx=t$ limits now become $0$ for $x=1$ and $1$ for $x=e$

$e^{t}=x$

$e^{t}.dt=dx$

by substituting the value value, we get

$\displaystyle \int^{1}_{0} e^{t} (\frac{(t-1)^2}{(t^2+1)^2}).dt$

$\displaystyle \int^{1}_{0} e^{t} (\frac{t^2-2t+1}{(t^2+1)^2}).dt$

$\displaystyle \int^{1}_{0} e^{t} (\frac{1}{t^2+1}-\frac{2t}{(t^2+1)^2}).dt$

As per general property $\displaystyle \int e^{x}(f(x)+f^{\prime}(x)).dx=e^{x}f(x)+c$

$\frac{1}{t^2+1}~is ~f(t)$ and $\frac{-2t}{(t^2+1)^2}~is~f^{\prime}(t)$

hence the result of integral is $\large{\frac{e^{t}}{t^2+1}}$

On applying limits

$\frac{e}{2}-1$

=> $\boxed{0.3591}$