What is the integral of 6/(sqrt(x^5)) ?

The integral of 6/(sqrt(x^5)) is -4/(x^{3/2)}+C

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integral of 6/(sqrt(x^5))

\int \frac{6}{x ^{5}} d x

- \frac{4}{x ^{\frac{3}{2}}} + C

Solution steps

\int \frac{6}{x ^{5}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{x ^{5}} d x

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

assuming

x \geq 0

= 6 \cdot \int \frac{1}{x ^{\frac{5}{2}}} d x

Apply the Power Rule

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

= 6 (- \frac{2}{3 x ^{\frac{3}{2}}})

Simplify

6 (- \frac{2}{3 x ^{\frac{3}{2}}}) : - \frac{4}{x ^{\frac{3}{2}}}

= - \frac{4}{x ^{\frac{3}{2}}}

Add a constant to the solution

= - \frac{4}{x ^{\frac{3}{2}}} + C

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What is the integral of 6/(sqrt(x^5)) ?
The integral of 6/(sqrt(x^5)) is -4/(x^{3/2)}+C