What is the integral from-7 to 2 of 1/((x-1)^{5/3)} ?

The integral from-7 to 2 of 1/((x-1)^{5/3)} is diverges

integral from-7 to 2 of 1/((x-1)^{5/3)}

\int_{- 7}^{2} \frac{1}{( x - 1 ) ^{\frac{5}{3}}} d x

diverges

Solution steps

\int_{- 7}^{2} \frac{1}{( x - 1 ) ^{\frac{5}{3}}} 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_{- 7}^{1} \frac{1}{( x - 1 ) ^{\frac{5}{3}}} d x + \int_{1}^{2} \frac{1}{( x - 1 ) ^{\frac{5}{3}}} d x

\int_{- 7}^{1} \frac{1}{( x - 1 ) ^{\frac{5}{3}}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 2}^{3} x^{- 3} d x

\int_{0}^{1} 15 e^{3 x} d x

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

\int_{1}^{4} (t + \frac{1}{t})^{2} d t

\int_{0}^{4} x \cdot \frac{x}{8} d x

What is the integral from-7 to 2 of 1/((x-1)^{5/3)} ?
The integral from-7 to 2 of 1/((x-1)^{5/3)} is diverges