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

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

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integral of 1/(-2x)

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

- \frac{1}{2} ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

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

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

Simplify

= (- \frac{1}{2}) \cdot \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= (- \frac{1}{2}) ln ∣ x ∣

Simplify

= - \frac{1}{2} ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ + C

cos^{2} (25 t)

5 x - 2 x

cos (\frac{2}{x})

\int sin (4 x - 2) d x

h (x) = x cos (x) + 3 π