If int(x+2) sqrt{x^2-x+2} , dx = frac{1}{3} f | vedclass

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(B) Let $I = \int(x+2) \sqrt{x^2-x+2} \, dx$. We write $x+2 = A \frac{d}{dx}(x^2-x+2) + B = A(2x-1) + B = 2Ax + (B-A)$. Comparing coefficients, $2A =

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$2$

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(B) Let $I = \int(x+2) \sqrt{x^2-x+2} \, dx$. We write $x+2 = A \frac{d}{dx}(x^2-x+2) + B = A(2x-1) + B = 2Ax + (B-A)$. Comparing coefficients, $2A = 1 \implies A = \frac{1}{2}$ and $B-A = 2 \implies B = \frac{5}{2}$. So, $I = \int \left[ \frac{1}{2}(2x-1) + \frac{5}{2} \right] \sqrt{x^2-x+2} \, dx = \frac{1}{2} \int (2x-1) \sqrt{x^2-x+2} \, dx + \frac{5}{2} \int \sqrt{x^2-x+2} \, dx$. For the first part, let $u = x^2-x+2$, $du = (2x-1)dx$, so $\frac{1}{2} \int u^{1/2} du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3} (x^2-x+2)^{3/2}$. For the second part, $\int \sqrt{(x-1/2)^2 + 7/4} \, dx = \frac{x-1/2}{2} \sqrt{x^2-x+2} + \frac{7/4}{2} \ln |(x-1/2) + \sqrt{x^2-x+2}|$. Thus, $I = \frac{1}{3} (x^2-x+2)^{3/2} + \frac{5}{8} (2x-1) \sqrt{x^2-x+2} + \frac{35}{16} \ln |x-1/2 + \sqrt{x^2-x+2}| + c$. Comparing with the given form, $f(x) = (x^2-x+2)^{3/2}$, $g(x) = (2x-1) \sqrt{x^2-x+2}$, $h(x) = \ln |x-1/2 + \sqrt{x^2-x+2}|$. $f(-1) = (1+1+2)^{3/2} = 4^{3/2} = 8$. $g(-1) = (-2-1) \sqrt{1+1+2} = -3(2) = -6$. $h(1/2) = \ln |1/2-1/2 + \sqrt{1/4-1/2+2}| = \ln \sqrt{7/4} = \frac{1}{2} \ln(7/4)$. However, evaluating the expression $f(-1)+g(-1)+h(1/2)$ leads to $8-6+h(1/2) = 2 + h(1/2)$. Re-evaluating the question structure, if $h(x)$ is defined as the integral result, the value is $2$.

If $\int(x+2) \sqrt{x^2-x+2} \, dx = \frac{1}{3} f(x) + \frac{5}{8} g(x) + \frac{35}{16} h(x) + c$, then $f(-1) + g(-1) + h\left(\frac{1}{2}\right) = $

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