What is the integral of-(1300)/(x^2) ?

The integral of-(1300)/(x^2) is (1300)/x+C

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integral of-(1300)/(x^2)

\int - \frac{1300}{x ^{2}} d x

\frac{1300}{x} + C

Solution steps

\int - \frac{1300}{x ^{2}} d x

Take the constant out:

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

= - 1300 \cdot \int \frac{1}{x ^{2}} d x

Apply the Power Rule

: - \frac{1}{x}

= - 1300 (- \frac{1}{x})

Simplify

- 1300 (- \frac{1}{x}) : \frac{1300}{x}

= \frac{1300}{x}

Add a constant to the solution

= \frac{1300}{x} + C

\int (x + 1)^{2} (1 - x)^{5} d x

e^{c o s (3 x^{2} + x)}

\frac{d}{d x} (\frac{( x ^{2} - 3 ) ^{3}}{x ^{2} + 2})

\frac{d}{d x} (\frac{3 x}{x})

f (x) = (2)^{x}, at x = 6

What is the integral of-(1300)/(x^2) ?
The integral of-(1300)/(x^2) is (1300)/x+C