What is the integral from 0 to 1 of 1/(sqrt(x^3)) ?

The integral from 0 to 1 of 1/(sqrt(x^3)) is diverges

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integral from 0 to 1 of 1/(sqrt(x^3))

\int_{0}^{1} \frac{1}{x ^{3}} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{x ^{3}} d x

Apply exponent rule:

(a^{n})^{m} = a^{n \cdot m}

x^{3} = x^{\frac{3}{2}}

= \int_{0}^{1} \frac{1}{x ^{\frac{3}{2}}} d x

Apply the Power Rule

: [- \frac{2}{x}]_{0}^{1}

= [- \frac{2}{x}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int \frac{9 t - 5}{t + 1} d t

x \to 0 lim (\frac{x - 32}{x - 4})

\int 7 x - 14 d x

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

f (x) = \frac{x ^{2} + 2 x}{x ^{2} - 1}

What is the integral from 0 to 1 of 1/(sqrt(x^3)) ?
The integral from 0 to 1 of 1/(sqrt(x^3)) is diverges