What is the integral from 5 to-5 of 1/(|x|^{2/3)} ?

The integral from 5 to-5 of 1/(|x|^{2/3)} is -6*5^{1/3}

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integral from 5 to-5 of 1/(|x|^{2/3)}

\int_{5}^{- 5} \frac{1}{∣ x ∣ ^{\frac{2}{3}}} d x

- 6 \cdot 5^{\frac{1}{3}}

Solution steps

\int_{5}^{- 5} \frac{1}{∣ x ∣ ^{\frac{2}{3}}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 5}^{5} \frac{1}{∣ x ∣ ^{\frac{2}{3}}} d x

Eliminate Absolutes

= - (\int_{- 5}^{0} \frac{1}{( - x ) ^{\frac{2}{3}}} d x + \int_{0}^{5} \frac{1}{x ^{\frac{2}{3}}} d x)

\int_{- 5}^{0} \frac{1}{( - x ) ^{\frac{2}{3}}} d x = - 3 (- 1)^{\frac{1}{3}} \cdot 5^{\frac{1}{3}}

\int_{0}^{5} \frac{1}{x ^{\frac{2}{3}}} d x = 3 \cdot 5^{\frac{1}{3}}

= - (- 3 (- 1)^{\frac{1}{3}} \cdot 5^{\frac{1}{3}} + 3 \cdot 5^{\frac{1}{3}})

Simplify

- (- 3 (- 1)^{\frac{1}{3}} \cdot 5^{\frac{1}{3}} + 3 \cdot 5^{\frac{1}{3}}) : - 6 \cdot 5^{\frac{1}{3}}

= - 6 \cdot 5^{\frac{1}{3}}

\int (1 + \frac{1}{x}) (\frac{1}{x ^{2}}) d x

y^{^{'}} = 9.8 - \frac{k y}{7}, y (0) = 94

e^{4 s i n (x)}

\frac{d}{d x} ((1 + 2 cos (x)) sin (x))

\frac{\partial}{\partial y} (sin (y) cos (x))

What is the integral from 5 to-5 of 1/(|x|^{2/3)} ?
The integral from 5 to-5 of 1/(|x|^{2/3)} is -6*5^{1/3}