What is the integral from 0 to 2 of 1/((x-1)^{2/5)} ?

The integral from 0 to 2 of 1/((x-1)^{2/5)} is 10/3

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integral from 0 to 2 of 1/((x-1)^{2/5)}

\int_{0}^{2} \frac{1}{( x - 1 ) ^{\frac{2}{5}}} d x

\frac{10}{3}

Decimal Notation

3.33333 \dots

Solution steps

\int_{0}^{2} \frac{1}{( x - 1 ) ^{\frac{2}{5}}} 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_{0}^{1} \frac{1}{( x - 1 ) ^{\frac{2}{5}}} d x + \int_{1}^{2} \frac{1}{( x - 1 ) ^{\frac{2}{5}}} d x

\int_{0}^{1} \frac{1}{( x - 1 ) ^{\frac{2}{5}}} d x = - \frac{5 ( - 1 ) ^{\frac{3}{5}}}{3}

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

= - \frac{5 ( - 1 ) ^{\frac{3}{5}}}{3} + \frac{5}{3}

Simplify

- \frac{5 ( - 1 ) ^{\frac{3}{5}}}{3} + \frac{5}{3} : \frac{10}{3}

= \frac{10}{3}

\int_{6}^{13} \frac{1}{( x - 8 ) ^{2}} d x

x y^{^{'}} - y = \frac{3}{4} x ln (x)

\frac{\partial}{\partial y} (2 x y)

\frac{d y}{d x}, x y = 24

What is the integral from 0 to 2 of 1/((x-1)^{2/5)} ?
The integral from 0 to 2 of 1/((x-1)^{2/5)} is 10/3