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

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}}

d er i v a t i v e \frac{1}{3}

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

a re a y = x 9 - x^{2}, y = 0, x = 0, x = 3

\frac{d}{d x} (arccos (6 x))

\int x^{3} \cdot ln (x) d 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}