What is the integral from 1 to infinity of x^{-1/5} ?

The integral from 1 to infinity of x^{-1/5} is diverges

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integral from 1 to infinity of x^{-1/5}

\int_{1}^{\infty} x^{- \frac{1}{5}} d x

diverges

Solution steps

\int_{1}^{\infty} x^{- \frac{1}{5}} d x

Compute the indefinite integral:

\int x^{- \frac{1}{5}} d x = \frac{5}{4} x^{\frac{4}{5}} + C

Compute the boundaries:

\int_{1}^{\infty} x^{- \frac{1}{5}} d x =

diverges

= diverges

x \to 1 + lim (\frac{x + 1}{x ^{2} - 1})

\int 2 sin (x) + 3 cos (x) d x

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 13

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

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

What is the integral from 1 to infinity of x^{-1/5} ?
The integral from 1 to infinity of x^{-1/5} is diverges