What is the integral from-2 to 7 of 1/((x+1)^{2/3)} ?

The integral from-2 to 7 of 1/((x+1)^{2/3)} is 9

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integral from-2 to 7 of 1/((x+1)^{2/3)}

\int_{- 2}^{7} \frac{1}{( x + 1 ) ^{\frac{2}{3}}} d x

9

Solution steps

\int_{- 2}^{7} \frac{1}{( x + 1 ) ^{\frac{2}{3}}} d x

Find undefined (singularity) points:

x = - 1

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_{- 2}^{- 1} \frac{1}{( x + 1 ) ^{\frac{2}{3}}} d x + \int_{- 1}^{7} \frac{1}{( x + 1 ) ^{\frac{2}{3}}} d x

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

\int_{- 1}^{7} \frac{1}{( x + 1 ) ^{\frac{2}{3}}} d x = 6

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

Simplify

- 3 (- 1)^{\frac{1}{3}} + 6 : 9

= 9

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What is the integral from-2 to 7 of 1/((x+1)^{2/3)} ?
The integral from-2 to 7 of 1/((x+1)^{2/3)} is 9