What is the integral of-(3000)/(x^2) ?

The integral of-(3000)/(x^2) is (3000)/x+C

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integral of-(3000)/(x^2)

\int - \frac{3000}{x ^{2}} d x

\frac{3000}{x} + C

Solution steps

\int - \frac{3000}{x ^{2}} d x

Take the constant out:

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

= - 3000 \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

: - \frac{1}{x}

= - 3000 (- \frac{1}{x})

Simplify

- 3000 (- \frac{1}{x}) : \frac{3000}{x}

= \frac{3000}{x}

Add a constant to the solution

= \frac{3000}{x} + C

\int \frac{10 x}{4 x ^{2} - 4} d x

\frac{( s + 3 )}{( s + 7 )}

y^{^{'}} + \frac{2 y}{t} = \frac{2 cos ( t ^{2} )}{t}

\int x \cdot ln (x + 1) d x

3 sin (3 x)

What is the integral of-(3000)/(x^2) ?
The integral of-(3000)/(x^2) is (3000)/x+C