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

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

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integral of 1/(-x)

integral

\frac{1}{- 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

in t e g r a l \frac{( y + 1 )}{( y - 1 )}

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

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

in t e g r a l sin (2 x + 3)

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