Answer
$$\int^{8}_1\frac{2+t}{\sqrt[3]{t^2}}dt=\frac{69}{4}$$
Work Step by Step
$$A=\int^{8}_1\frac{2+t}{\sqrt[3]{t^2}}dt$$ $$A=\int^{8}_1(2+t)t^{-2/3}dt$$ $$A=\int^{8}_1(2t^{-2/3}+t^{1/3})dt$$
According to Table 1, we have $$\int(x^n)dx=\frac{x^{n+1}}{n+1}+C(n\ne-1)$$ Therefore, $$A=(\frac{2t^{1/3}}{\frac{1}{3}}+\frac{t^{4/3}}{\frac{4}{3}})\Bigg]^{8}_1$$ $$A=(6\sqrt[3]t+\frac{3\sqrt[3]{t^4}}{4})\Bigg]^{8}_1$$ $$A=(6\sqrt[3]8+\frac{3\sqrt[3]{8^4}}{4})-(6\sqrt[3]1+\frac{3\sqrt[3]{1^4}}{4})$$ $$A=(6\times2+\frac{3\times16}{4})-(6+\frac{3}{4})$$ $$A=(12+12)-(\frac{27}{4})$$ $$A=\frac{69}{4}$$
