Question

,

Answer 1

the first derivative is the integrand so \[y'=\frac{1}{1+3x+x^2}\]

Answer 2

take the derivative of that one to test concavity

Answer 3

(-2 x-3)/(x^2+3 x+1)^2 ?

Answer 4

yes looks reasonable

Answer 5

so it the second derivative is positive, it is concave up
solve \[-2x-3>0\]to find that interval

Answer 6

im confused what do I need to do?

Answer 7

take the second derivative
you did that already
find where it is positive

Answer 8

the denominator is always positive, because it is a square
so all that is left to do is solve \[-2x-3>0\] in two simple steps

Answer 9

x<-3/2?