What is the integral of 1/(sqrt(x^5)) ?

The integral of 1/(sqrt(x^5)) is -2/(3x^{3/2)}+C

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integral of 1/(sqrt(x^5))

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

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

Solution steps

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

x^{5} = x^{\frac{5}{2}},

assuming

x \geq 0

= \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

x \to - 4 lim (\frac{5}{x + 4})

\frac{d u}{1 + sin ( u )} = d x

5 t \cdot \frac{d y}{d t} - 5 y = t

f (x) = x^{x^{2}}

f (x) = 9 x + 4, (5, 7)

What is the integral of 1/(sqrt(x^5)) ?
The integral of 1/(sqrt(x^5)) is -2/(3x^{3/2)}+C