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{d y}{d x} = \frac{1}{2} y

y = \frac{4 x ^{2} + 4 x + 2}{x}

\int_{0}^{3} \frac{1}{3 - x} d x

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 e^{- 2 t}

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