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

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

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

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

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

Decimal Notation

1.28589 \dots

Solution steps

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

Find undefined (singularity) points:

x = 3

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

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

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

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

Simplify

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

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

(\frac{y ^{^{'}} - e ^{- t} + 5}{y}) = - 5, y (0) = - 2

\int \frac{5}{9 - ( x - 3 ) ^{2}} d x

y = \frac{2}{x - 4}

\int \frac{1}{k - x} d x

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