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

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

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Find the Antiderivative f(x)=1/(x^2)

integral

f (x) = \frac{1}{( x ^{2} )}

- \frac{1}{x} + C

Solution steps

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

Apply the Power Rule

: - \frac{1}{x}

= - \frac{1}{x}

Add a constant to the solution

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

e^{2} x

y = sec (x)

- \frac{d}{d x} f (x) = arcsec (x)

- \frac{d}{d x} (\frac{x}{4})

- \frac{d}{d x} (\frac{100}{x})

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