Question

Find the interval on which the curve of ∫ 6/(1+2t+t^2) where b=0 and a=x is concave up.
Heres my work but im not sure at all

g '(x) = 6/(1+2x+x^2)
g ''(x) = -(2+2x)/[(1+2x+x^2)^2]

upward when g ''(x) > 0.
- (2+2x) > 0
x < -1

Answer 1

I was mistaken, it is b=x and a=0

Answer 2

-(x+1)>0
x+1 < 0
x <-1

I seem to agree but try
http://www.wolframalpha.com/input/?i=+integrate+6%2F%281%2B2x+%2B+x^2%29

Which agrees

Answer 3

I assumed you were integrating from 0 to x