What is the integral of (-1000)/(x^2) ?

The integral of (-1000)/(x^2) is (1000)/x+C

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integral of (-1000)/(x^2)

\int \frac{- 1000}{x ^{2}} d x

\frac{1000}{x} + C

Solution steps

\int \frac{- 1000}{x ^{2}} d x

Take the constant out:

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

= - 1000 \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

: - \frac{1}{x}

= - 1000 (- \frac{1}{x})

Simplify

- 1000 (- \frac{1}{x}) : \frac{1000}{x}

= \frac{1000}{x}

Add a constant to the solution

= \frac{1000}{x} + C

\frac{s ^{2} + s}{s ^{2} + s + 1}

y^{^{''}} - 3 y^{^{'}} + 7 y = x e^{x}

\int \frac{8}{12 - x ^{2} - 4 x} d x

3 x^{2}

\int (2 x + 1)^{7} d x

What is the integral of (-1000)/(x^2) ?
The integral of (-1000)/(x^2) is (1000)/x+C