If f(x) = begin{cases} frac{6 x^2 + 1}{4 x^3 | vedclass

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(A) To evaluate $\int_0^2 f(x) dx$,we split the integral at $x = 1$: $\int_0^2 f(x) dx = \int_0^1 \frac{6 x^2 + 1}{4 x^3 + 2 x + 3} dx + \int_1^2 (x^2

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$\frac{1}{2} \log 3 + \frac{10}{3}$

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(A) To evaluate $\int_0^2 f(x) dx$,we split the integral at $x = 1$: $\int_0^2 f(x) dx = \int_0^1 \frac{6 x^2 + 1}{4 x^3 + 2 x + 3} dx + \int_1^2 (x^2 + 1) dx$ For the first integral,let $u = 4 x^3 + 2 x + 3$,then $du = (12 x^2 + 2) dx = 2(6 x^2 + 1) dx$. Thus,$\int_0^1 \frac{6 x^2 + 1}{4 x^3 + 2 x + 3} dx = \frac{1}{2} \int_{u(0)}^{u(1)} \frac{1}{u} du = \frac{1}{2} [\log |u|]_3^9 = \frac{1}{2} (\log 9 - \log 3) = \frac{1}{2} \log 3$. For the second integral,$\int_1^2 (x^2 + 1) dx = [\frac{x^3}{3} + x]_1^2 = (\frac{8}{3} + 2) - (\frac{1}{3} + 1) = \frac{14}{3} - \frac{4}{3} = \frac{10}{3}$. Adding both parts,$\int_0^2 f(x) dx = \frac{1}{2} \log 3 + \frac{10}{3}$.

If $f(x) = \begin{cases} \frac{6 x^2 + 1}{4 x^3 + 2 x + 3}, & 0 < x < 1 \\ x^2 + 1, & 1 \le x \le 2 \end{cases}$ then $\int_0^2 f(x) dx =$

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