What is the integral of-(1600)/(x^2) ?

The integral of-(1600)/(x^2) is (1600)/x+C

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integral of-(1600)/(x^2)

\int - \frac{1600}{x ^{2}} d x

\frac{1600}{x} + C

Solution steps

\int - \frac{1600}{x ^{2}} d x

Take the constant out:

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

= - 1600 \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

= - 1600 (- \frac{1}{x})

Simplify

- 1600 (- \frac{1}{x}) : \frac{1600}{x}

= \frac{1600}{x}

Add a constant to the solution

= \frac{1600}{x} + C

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What is the integral of-(1600)/(x^2) ?
The integral of-(1600)/(x^2) is (1600)/x+C