$\int_0^8 (x^{\frac{2}{3}} + 1) dx$ is equal to

The integral can be split into two parts:

$\int_0^8 (x^{\frac{2}{3}} + 1) dx = \int_0^8 x^{\frac{2}{3}} dx + \int_0^8 1 dx$

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Using the power rule with $n = \frac{2}{3}$:

$n + 1 = \frac{2}{3} + 1 = \frac{5}{3}$

$\int x^{\frac{2}{3}} dx = \frac{x^{\frac{5}{3}}}{\frac{5}{3}} = \frac{3}{5}x^{\frac{5}{3}}$

For the constant:

$\int 1 dx = x$

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Combining the results:

$\int_0^8 (x^{\frac{2}{3}} + 1) dx = \left[\frac{3}{5}x^{\frac{5}{3}} + x\right]_0^8$

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Evaluating $8^{\frac{5}{3}}$:

$8^{\frac{5}{3}} = (8^{\frac{1}{3}})^5$

$8^{\frac{1}{3}} = 2$

$(2)^5 = 32$

At $x = 8$:

$\frac{3}{5}(32) + 8 = \frac{96}{5} + \frac{40}{5} = \frac{136}{5}$

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At $x = 0$:

$\frac{3}{5}(0)^{\frac{5}{3}} + 0 = 0$

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The final result:

$\frac{136}{5} - 0 = \frac{136}{5}$

Therefore, $\int_0^8 (x^{\frac{2}{3}} + 1) dx = \dfrac{136}{5}$