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

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

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

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

\frac{1}{2} ln ∣ x - 2 ∣ + C

Solution steps

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

Apply u-substitution

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

= \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 = x - 2

= \frac{1}{2} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x - 2 ∣ + C

\int_{0}^{1} x^{2} e^{x^{3} - 1} d x

\frac{\partial}{\partial x} (x^{2} + y^{3} + w^{4})

y = ∣ 3 x ∣, y = x^{2} - 4

x \to \infty lim (34)

f (x) = \frac{2 x}{x ^{2} + 1}

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