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

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

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

integral

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

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

Solution steps

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

Simplify

\frac{x ^{- 1}}{x ^{2}} : \frac{1}{x ^{3}}

= \int \frac{1}{x ^{3}} d x

Apply the Power Rule

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

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

Add a constant to the solution

= - \frac{1}{2 x ^{2}} + C

- sin^{2} (x) + tan^{2} (x)

\frac{x ^{2}}{( x ^{2} + 1 )}

\frac{1}{( x + 2 )}

\frac{( e ^{x} ) ^{2}}{2}

\frac{1}{( x ^{2} - 1 ) ^{3}}

What is the integral of (x^{-1})/(x^2) ?
The integral of (x^{-1})/(x^2) is -1/(2x^2)+C