What is the integral of 6/(sqrt(x^3)) ?

The integral of 6/(sqrt(x^3)) is -(12)/(sqrt(x))+C

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integral of 6/(sqrt(x^3))

\int \frac{6}{x ^{3}} d x

- \frac{12}{x} + C

Solution steps

\int \frac{6}{x ^{3}} 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 ^{3}} d x

Apply exponent rule:

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

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

= 6 \cdot \int \frac{1}{x ^{\frac{3}{2}}} d x

Apply the Power Rule

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

Simplify

6 (- \frac{2}{x}) : - \frac{12}{x}

= - \frac{12}{x}

Add a constant to the solution

= - \frac{12}{x} + C

What is the integral of 6/(sqrt(x^3)) ?
The integral of 6/(sqrt(x^3)) is -(12)/(sqrt(x))+C