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

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

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

\int_{0}^{6} \frac{1}{( x - 4 ) ^{\frac{2}{3}}} d x

3 \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

Decimal Notation

8.54196 \dots

Solution steps

\int_{0}^{6} \frac{1}{( x - 4 ) ^{\frac{2}{3}}} d x

Find undefined (singularity) points:

x = 4

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}^{4} \frac{1}{( x - 4 ) ^{\frac{2}{3}}} d x + \int_{4}^{6} \frac{1}{( x - 4 ) ^{\frac{2}{3}}} d x

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

\int_{4}^{6} \frac{1}{( x - 4 ) ^{\frac{2}{3}}} d x = 3 \cdot 2^{\frac{1}{3}}

= - 3 (- 1)^{\frac{1}{3}} \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

Simplify

- 3 (- 1)^{\frac{1}{3}} \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}} : 3 \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

= 3 \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}

\int 9 cos (x) d x

\int \frac{5 x ^{4} + 4}{x ^{5} + 4 x} d x

\int_{0}^{\frac{π}{4}} sec^{2} (t) d t

\frac{d y}{d x} = \frac{1 - 2 x}{4 x - x ^{2}}

y + 6 y^{3}

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