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

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

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

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

6

Solution steps

\int_{- 1}^{1} \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_{- 1}^{0} \frac{1}{x ^{\frac{2}{3}}} d x + \int_{0}^{1} \frac{1}{x ^{\frac{2}{3}}} d x

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

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

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

Simplify

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

= 6

\frac{d}{d x} ((\frac{x + 1}{x})^{3})

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 16})

\int x^{9} ln (x) d x

f (x) = (13 x - 1)^{- \frac{1}{3}}, at x = 5

\frac{t}{( t - 9 ) ^{2}}

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