What is the integral of-1/x ?

The integral of-1/x is -ln|x|+C

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integral of-1/x

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

- ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= - ln ∣ x ∣

Add a constant to the solution

= - ln ∣ x ∣ + C

\frac{\partial}{\partial x} (x^{2} + y^{3})

\frac{d}{d x} (y - \frac{1}{( x + 3 )})

\frac{d y}{d x} - 7 y = e^{x}

\frac{\partial}{\partial x} (\frac{( u - v )}{u + v})

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