What is the integral of 1/(x^2sqrt(x)) ?

The integral of 1/(x^2sqrt(x)) is -2/(3x^{3/2)}+C

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integral of 1/(x^2sqrt(x))

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

- \frac{2}{3 x ^{\frac{3}{2}}} + C

Solution steps

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

Simplify

x^{2} x : x^{\frac{5}{2}}

= \int \frac{1}{x ^{\frac{5}{2}}} d x

Apply the Power Rule

: - \frac{2}{3 x ^{\frac{3}{2}}}

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

Add a constant to the solution

= - \frac{2}{3 x ^{\frac{3}{2}}} + C

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

What is the integral of 1/(x^2sqrt(x)) ?
The integral of 1/(x^2sqrt(x)) is -2/(3x^{3/2)}+C