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

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

-(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

I assumed you were integrating from 0 to x

To determine the interval on which the curve is concave up, we need to find where the second derivative of the function is positive.

First, let's find the second derivative of the function:
g''(x) = -(2+2x)/[(1+2x+x^2)^2]

Next, we need to set g''(x) > 0 to find the values of x where the curve is concave up.

-(2+2x) > 0

To solve this inequality, we can multiply both sides by -1, which reverses the inequality:

2+2x < 0

Subtracting 2 from both sides:

2x < -2

Dividing both sides by 2:

x < -1

Therefore, the curve is concave up for x values less than -1.

To summarize, the interval on which the curve is concave up is (-∞, -1).