Question 1156000
i believe i have the answer you're looking for.
as you indicated, the maximum value is 1.386 rounded to 3 decimal places.  this occurred at x = 1.
this i was able to determine by graphing the equation and looking for the maximum value.
using my calculator, i was able to determine that the maximum value was 1.386294361 rounded to the number of digits the calculator was able to display.
it was evident from the graph that x = 1 was the highest point in the interval from -1 to 1.
using the same graph, i was able to determine that the minimum value was.56 when x = .5.
i then looked up the first derivative of ln(x^2 + x + 2).  
according to the derivative calculator i found on the web, it was (2x + 1) / (x^2 + x + 2).
i set this equal to 0 and solved for x to get x = -1/2.
when x = -1/2, i used a calculator on the original equation to get the value of .5596157879 rounded to the number of decimal digits the calculator was able to display.
i used information that i got from the web and used a graphing calculator to determine what the answer should be.
since i didn't really know how to find the derivative of the ln function, i did a little research on the web and came up with the following reference that you might find useful.
that reference is <a href = "http://www.ltcconline.net/greenl/courses/116/explog/logderivative.htm" target = "_blank">http://www.ltcconline.net/greenl/courses/116/explog/logderivative.htm</a>
it's a little dificult to follow with all the dy/dx jargon, but the general idea is that the derivative of ln(x) = 1/x and the derivative of an expression within the parentheses of ln(x), such as ln(3x-4) involves the chain rule and winds up being the derivative of 3 divided by 3x-4.
if you extrapolate from this example, you see that the derivative of ln(x^2 + x + 2) is the derivative of x^2 + x + 2 divided by x^2 + x + 2.
this winds up being 2x + 1 divided by x^2 + x + 2.
here's what the graph looks like.
<img src = "http://theo.x10hosting.com/2020/041401.jpg" alt="$$$" >
here's what the output of the derivative calculator looks like.
<img src = "http://theo.x10hosting.com/2020/041402.jpg" alt="$$$" >
the calculator can be found at <a href = "https://www.desmos.com/calculator" target = "_blank">https://www.desmos.com/calculator</a>
the derivative calculator can be found at <a href = "https://www.derivative-calculator.net/" target = "_blank">https://www.derivative-calculator.net/</a>