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}}

Decimal Notation

10.25985 \dots

Solution steps

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

Find undefined (singularity) points:

x = 0

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \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}}

- arcsin (\frac{1}{4} x) + \frac{x}{16 - x ^{2}} + C

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

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

\int_{- 2}^{4} x - 9 d x

f (x) = 6 (x - 2)

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}