We can observe that the function at the right hand is positive [math]\forall x[/math], but the function at the left hand is positive only in the interval [math]] 1/e , 1/\sqrt{e} [[/math].

So the solution is in the previous interval:

[math]\frac{1}{e} < x < \frac{1}{\sqrt{2}}[/math]

Now you rewrite your equation in the following mode:

[math]e^{-\frac{x+1}{2x+1}} + \frac{1+\ln x}{1 + 2 \ln x} = g(x)[/math]

and apply the Bisection method - Wikipedia to

[math]g(x) = 0[/math]

You should find one solution.