How do you evaluate the integral $\int \frac{1}{{(x - 2)}^{\frac{2}{3}}} d x$ $int 1/(x-2)^(2/3)dx$ from 1 to 4?

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How do you evaluate the integral $\int \frac{1}{{(x - 2)}^{\frac{2}{3}}} d x$ $int 1/(x-2)^(2/3)dx$ from 1 to 4?

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Answer 1 · 2016-12-30T13:49:05

$\int_{1}^{4} \frac{d x}{{(x - 2)}^{\frac{2}{3}}} = 3 (\sqrt[3]{2} + 1)$ $int_1^4 (dx)/(x-2)^(2/3) = 3(root(3)2+1)$

Explanation:

$\int_{1}^{4} \frac{d x}{{(x - 2)}^{\frac{2}{3}}} = {[3 {(x - 2)}^{\frac{1}{3}}]}_{1}^{\frac{1}{3}} = 3 (2^{\frac{1}{3}} - {(- 1)}^{\frac{1}{3}}) = 3 (\sqrt[3]{2} + 1)$ $int_1^4 (dx)/(x-2)^(2/3) = [3(x-2)^(1/3)]_1^4 = 3(2^(1/3)-(-1)^(1/3))= 3(root(3)2+1)$

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How do you evaluate the integral $\int \frac{1}{{(x - 2)}^{\frac{2}{3}}} d x$ $int 1/(x-2)^(2/3)dx$ from 1 to 4?

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How do you evaluate the integral $\int \frac{1}{{(x - 2)}^{\frac{2}{3}}} d x$ $int 1/(x-2)^(2/3)dx$ from 1 to 4?