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

The integral from-1 to 1 of sqrt(1+x^{-2/3)} is 0

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

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

0

Solution steps

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

\int_{- 1}^{0} 1 + x^{- \frac{2}{3}} d x = - 22 + 1

\int_{0}^{1} 1 + x^{- \frac{2}{3}} d x = 22 - 1

= - 22 + 1 + 22 - 1

Simplify

= 0

t \to 1 lim (\frac{1}{t ^{2} - 1})

(f (t) + g (t))^{'}

\int x^{2} + 6 d x

\int x + cos (x) d x

\frac{d}{d x} (2 x sec (x))

What is the integral from-1 to 1 of sqrt(1+x^{-2/3)} ?
The integral from-1 to 1 of sqrt(1+x^{-2/3)} is 0