What is the integral of (6/(x^4)) ?

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

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integral of (6/(x^4))

\int (\frac{6}{x ^{4}}) d x

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

Solution steps

\int \frac{6}{x ^{4}} 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 ^{4}} d x

Apply the Power Rule

: - \frac{1}{3 x ^{3}}

= 6 (- \frac{1}{3 x ^{3}})

Simplify

6 (- \frac{1}{3 x ^{3}}) : - \frac{2}{x ^{3}}

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

Add a constant to the solution

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

\frac{d}{d x} (e^{2 x} 2)

\frac{\partial}{\partial x} (x y cos (yz))

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

a^{2} x^{2} + b^{2}