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

= - 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

d er i v a t i v e 2^{t}

\int \frac{8 x - 2}{x ^{2} - 1} d x

\int e^{9 x} d x

a re a y = x^{3} - x, y = 3 x, - 2, 2

\frac{d}{d x} ((1 + 3 x^{2}) (x - x^{2}))