What is the integral of (-(4000)/(x^2)) ?

The integral of (-(4000)/(x^2)) is (4000)/x+C

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integral of (-(4000)/(x^2))

\int (- \frac{4000}{x ^{2}}) d x

\frac{4000}{x} + C

Solution steps

\int - \frac{4000}{x ^{2}} d x

Take the constant out:

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

= - 4000 \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

: - \frac{1}{x}

= - 4000 (- \frac{1}{x})

Simplify

- 4000 (- \frac{1}{x}) : \frac{4000}{x}

= \frac{4000}{x}

Add a constant to the solution

= \frac{4000}{x} + C

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What is the integral of (-(4000)/(x^2)) ?
The integral of (-(4000)/(x^2)) is (4000)/x+C