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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

: - \frac{1}{x}

= - 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

\int sin (3 x) \cdot cos (2 x) d x

\int x + \frac{1}{5 x} d x

\int (3 - 3 x^{2}) d x

\int e^{(5 x)} d x

\int \frac{1}{x 81 x ^{2} - 1} d x

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