Answer:
Mark as brainliest if helped Thanks!
Step-by-step explanation:
funtion is:
[tex]y = \frac{x^2 -1}{2x + 1}[/tex]
therefore [tex]\\\frac{dy}{dx}\\[/tex] is
[tex]\frac{d\frac{x^2 -1}{2x + 1}}{dx}[/tex]
[tex]\frac{\frac{x^2 -1}{2x + 1}}{x}[/tex]
[tex]\frac{x^2 -1}{(2x+1) x}[/tex]
Now
[tex]\int\limits^2_0 {\frac{x^2 +x +1 }{4x^2 +4 + 1} } \, dx[/tex]
(Calculation of this it too long and it is hard to type on brainly)
here is the answer though....
= 4/5
=0.8
