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

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

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integral of 1/(1-2v)

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

- \frac{1}{2} ln ∣ 1 - 2 v ∣ + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{2 u} d u

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 - 2 v

= - \frac{1}{2} ln ∣ 1 - 2 v ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ 1 - 2 v ∣ + C

e^{2 x}, 5

\frac{d y}{d x} = 2 x y^{2}

\frac{d}{d x} (169 - x^{2})

\frac{\partial}{\partial x} (ln (2 x + y))