What is the integral of (-1)/(x^2) ?

The integral of (-1)/(x^2) is 1/x+C

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integral of (-1)/(x^2)

integral

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

\frac{1}{x} + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

Simplify

= \frac{1}{x}

Add a constant to the solution

= \frac{1}{x} + C

in t e g r a l 1 + x

in t e g r a l 1 - cos^{2} (x)

in t e g r a l \frac{x}{( x ^{2} + 1 )}

in t e g r a l \frac{1}{( 1 - x ^{2} )}

in t e g r a l \frac{x ^{3}}{( x + 1 ) ^{2}}