What is the integral from 0 to 4 of 1/((x-2)^{2/3)} ?

The integral from 0 to 4 of 1/((x-2)^{2/3)} is 6*2^{1/3}

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integral from 0 to 4 of 1/((x-2)^{2/3)}

\int_{0}^{4} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x

6 \cdot 2^{\frac{1}{3}}

Decimal

7.55952 \dots

Solution steps

\int_{0}^{4} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x

Find undefined (singularity) points:

x = 2

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_{0}^{2} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x + \int_{2}^{4} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x

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

\int_{2}^{4} \frac{1}{( x - 2 ) ^{\frac{2}{3}}} d x = 3 \cdot 2^{\frac{1}{3}}

= - 3 (- 1)^{\frac{1}{3}} \cdot 2^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

Simplify

- 3 (- 1)^{\frac{1}{3}} \cdot 2^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}} : 6 \cdot 2^{\frac{1}{3}}

= 6 \cdot 2^{\frac{1}{3}}

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\int_{- 4}^{1} - x^{2} - 3 x + 4 d x

What is the integral from 0 to 4 of 1/((x-2)^{2/3)} ?
The integral from 0 to 4 of 1/((x-2)^{2/3)} is 6*2^{1/3}