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

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

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integral of 1/(1-1/2 x)

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

- 2 ln 1 - \frac{1}{2} x + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 ln ∣ u ∣

Substitute back

u = 1 - \frac{1}{2} x

= - 2 ln 1 - \frac{1}{2} x

Add a constant to the solution

= - 2 ln 1 - \frac{1}{2} x + C

\frac{x}{6}

(x tan (\frac{y}{x}) + y) \cdot d x - x \cdot d y = 0

\int \frac{1}{x} - \frac{1}{x + 6} d x

\int y 2 + 10 y - 25 y^{2} d y

\frac{( x + 1 ) ^{2}}{( x - 1 ) ( x + 2 )}

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