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

\frac{d}{d x} (sin^{2} (x) cos^{3} (x))

x \to 0 lim (x sin (\frac{4}{x}))

y^{^{'}} + y + 2 cos (x) = 0

\int_{4}^{6} \frac{1}{x} d x

x \to - 1 lim (x^{3} - 4 x + 1)

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