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

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

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

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

diverges

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (e^{x^{2}} + x^{2})

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

y = ln (7 x^{2} + 9 y^{2})

\int ln (1 + 2 x) d x

\int (2 x y + x^{2} - 1) d y

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