What is the integral from 1 to 4 of 1/((x-3)^{1/5)} ?

The integral from 1 to 4 of 1/((x-3)^{1/5)} is -5/(2*2^{1/5)}+5/4

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integral from 1 to 4 of 1/((x-3)^{1/5)}

\int_{1}^{4} \frac{1}{( x - 3 ) ^{\frac{1}{5}}} d x

- \frac{5}{2 \cdot 2 ^{\frac{1}{5}}} + \frac{5}{4}

Decimal Notation

- 0.92637 \dots

Solution steps

\int_{1}^{4} \frac{1}{( x - 3 ) ^{\frac{1}{5}}} 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_{1}^{3} \frac{1}{( x - 3 ) ^{\frac{1}{5}}} d x + \int_{3}^{4} \frac{1}{( x - 3 ) ^{\frac{1}{5}}} d x

\int_{1}^{3} \frac{1}{( x - 3 ) ^{\frac{1}{5}}} d x = - \frac{5 ( - 1 ) ^{\frac{4}{5}}}{2 \cdot 2 ^{\frac{1}{5}}}

\int_{3}^{4} \frac{1}{( x - 3 ) ^{\frac{1}{5}}} d x = \frac{5}{4}

= - \frac{5 ( - 1 ) ^{\frac{4}{5}}}{2 \cdot 2 ^{\frac{1}{5}}} + \frac{5}{4}

5 (- 1)^{\frac{4}{5}} = 5

= - \frac{5}{2 \cdot 2 ^{\frac{1}{5}}} + \frac{5}{4}

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What is the integral from 1 to 4 of 1/((x-3)^{1/5)} ?
The integral from 1 to 4 of 1/((x-3)^{1/5)} is -5/(2*2^{1/5)}+5/4