91Ó°ÊÓ

We'll first find the derivative of \(I(x)\) with respect to \(x\), using the Fundamental Theorem of Calculus, which states that if \(F(x) = \int_{a}^{x} f(t) dt\), then \(F'(x) = f(x)\). So for our function, we get: \(I'(x) = \frac{2x + 1}{x^2 - 2x + 2}\).

To find the critical points of \(I'(x)\), we need to set the derivative equal to zero and solve for \(x\): \(\frac{2x + 1}{x^2 - 2x + 2} = 0\). However, we can see that the numerator is a linear function and the denominator is a quadratic function with a positive leading coefficient and no real roots (discriminant is negative). This means that the fraction can never be equal to zero. Thus, there are no critical points for \(I(x)\).

Since there are no critical points, we only need to evaluate \(I(x)\) at the interval's endpoints (-1 and 1) to find the least and greatest values: 1. \(I(-1) = \int_{0}^{-1} \frac{2t + 1}{t^2 - 2t + 2} dt\) and 2. \(I(1) = \int_{0}^{1} \frac{2t + 1}{t^2 - 2t + 2} dt\). We can find these values by evaluating the definite integrals.

To evaluate the definite integrals, we need to integrate the function \(\frac{2t + 1}{t^2 - 2t + 2}\). We notice that \(t^2 - 2t + 2 = (t-1)^2 + 1\). In order to integrate, we'll make a substitution: let \(u = t - 1\). Then \(t = u + 1\) and \(dt = du\). The function becomes: \(\frac{2(u+1) + 1}{u^2 + 1}du = \frac{2u + 3}{u^2 + 1}du\). Now, integrating with respect to \(u\), we get: \(I(u) = \int_{\text{new lower limit}}^{\text{new upper limit}} \frac{2u + 3}{u^2 + 1} du\). (1) Next, we need to calculate the new integration limits. When \(x = -1\), \(u = -2\). When \(x = 1\), \(u = 0\). Applying the new limits to the integral (1), we get: \(I(-2) = \int_{0}^{-2} \frac{2u + 3}{u^2 + 1} du\) and \(I(0) = \int_{0}^{0} \frac{2u + 3}{u^2 + 1} du\). Notice that the second integral is trivially zero since it has the same lower and upper limits. Now we just need to evaluate the first integral to find the least and greatest values of \(I(x)\) on the interval \([-1, 1]\). Evaluating the first integral, we can split it into two parts: \(I(-2) = \int_{0}^{-2} \frac{2u}{u^2 + 1} du + \int_{0}^{-2} \frac{3}{u^2 + 1} du\). Using basic integration techniques (integration by substitution and arctangent rule), we find: \(I(-2) = \left[-\frac{\ln(u^2+1)}{2}\right]_{0}^{-2} + \left[3\arctan(u)\right]_{0}^{-2}\). Evaluating the integrals, we get: \(I(-2) = -\frac{1}{2}\ln(5) + \frac{1}{2}\ln(1) + 3\arctan(-2) - 3\arctan(0) = -\frac{1}{2}\ln(5) + 3\arctan(-2)\).

Now that we have computed the values of \(I(-1)\) and \(I(1)\), we can conclude that the greatest and least values of the function on the interval \([-1,1]\) are: Greatest value: \(I(1) = 0\). Least value: \(I(-1) = -\frac{1}{2}\ln(5) + 3\arctan(-2)\).