What is the integral from 0 to 5 of 1/(x^{19/18)} ?

The integral from 0 to 5 of 1/(x^{19/18)} is diverges

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integral from 0 to 5 of 1/(x^{19/18)}

\int_{0}^{5} \frac{1}{x ^{\frac{19}{18}}} d x

diverges

Solution steps

\int_{0}^{5} \frac{1}{x ^{\frac{19}{18}}} d x

Apply the Power Rule

: [- \frac{18}{x ^{\frac{1}{18}}}]_{0}^{5}

= [- \frac{18}{x ^{\frac{1}{18}}}]_{0}^{5}

Compute the boundaries:

diverges

= diverges

f (x) = \frac{3}{x ^{2}}, (1, 3)

\frac{d}{d x} ((9 x - 6)^{2})

\int \frac{1}{10 + 4 x - x ^{2}} d x

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

x y^{^{'}} + 3 y = \frac{( 2 e ^{- 2 x} )}{x ^{2}}

What is the integral from 0 to 5 of 1/(x^{19/18)} ?
The integral from 0 to 5 of 1/(x^{19/18)} is diverges