Find the interval on which the curve of y equals the integral from 0 to x of 6 divided by the quantity 1 plus 2 times t plus t squared, dt is concave up.

To find the interval on which the curve of y = ∫(0 to x) (6/(1+2t+t^2)) dt is concave up, we need to determine the intervals where the second derivative of y is positive.

Let's start by finding the first derivative of y with respect to x. We can use the Fundamental Theorem of Calculus to differentiate the integral.

The given function is: y = ∫(0 to x) (6/(1+2t+t^2)) dt

To find its derivative, we need to apply the Chain Rule. Let's define a new variable, say u, as the upper limit of integration:
u = x

Now, we can rewrite the integral in terms of u:
y = ∫(0 to u) (6/(1+2t+t^2)) dt

Next, we differentiate y with respect to x by substituting back u in place of x and differentiating with respect to u:
dy/dx = d/dx [∫(0 to u) (6/(1+2t+t^2)) dt] = d/du [∫(0 to u) (6/(1+2t+t^2)) dt] × du/dx

Notice that the second term, du/dx, is just the derivative of the variable u with respect to x, which is 1.

So, we have:
dy/dx = d/du [∫(0 to u) (6/(1+2t+t^2)) dt] × 1

To differentiate the integral, we can apply Leibniz's Rule for Differentiating under the Integral Sign. According to this rule, if the integrand (6/(1+2t+t^2)) is continuous in both t and u, then:

dy/dx = (d/du ∫(0 to u) (6/(1+2t+t^2)) dt)

Let's find this derivative:
To apply the rule, let's define a new function, say F(t,u), as follows:
F(t,u) = ∫(0 to u) (6/(1+2t+t^2)) dt

Now, we need to differentiate F(t,u) with respect to u.

∂F/∂u = d/dx ∫(0 to u) (6/(1+2t+t^2)) dt

We will treat t as a constant since we are differentiating with respect to u. Applying the Fundamental Theorem of Calculus, we obtain:

∂F/∂u = (6/(1+2u+u^2)) × du/dx = (6/(1+2u+u^2))

Now, let's find the second derivative of y by differentiating dy/dx with respect to x:

d^2y/dx^2 = d/dx (dy/dx)

To do this, we apply the Chain Rule again:
d^2y/dx^2 = d/dx [(6/(1+2u+u^2)) × 1]

Now, we can simplify this expression:
d^2y/dx^2 = (d/du [(6/(1+2u+u^2)) × 1]) × du/dx

Using the result we got earlier, du/dx = 1, we have:
d^2y/dx^2 = (d/du [(6/(1+2u+u^2))]) × 1

To differentiate (6/(1+2u+u^2)) with respect to u, we use the quotient rule:
d^2y/dx^2 = ([(1+2u+u^2) * 0 - 6 * (2+2u)] / [(1+2u+u^2)^2]) × 1

Simplifying:
d^2y/dx^2 = (-12 - 12u) / (1+2u+u^2)^2

To find the intervals where the curve is concave up, we need to determine when the second derivative is positive. Let's set it greater than zero and solve for u:

(-12 - 12u) / (1+2u+u^2)^2 > 0

Since the denominator is always positive, the numerator must be negative for the inequality to hold.

-12 - 12u < 0

Simplifying:
12u > -12

Dividing by 12:
u > -1

Therefore, the interval where the curve is concave up is u > -1.

Converting back to the original variable x, we have: x > -1.

So, the interval on which the curve is concave up is x > -1.