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

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

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integral of 1/(2x-2)

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

\frac{1}{2} ln ∣ 2 x - 2 ∣ + C

Solution steps

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

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 = 2 x - 2

= \frac{1}{2} ln ∣ 2 x - 2 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x - 2 ∣ + C

\int 2 cos^{2} (4 x) sin (4 x) d x

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{\frac{1}{2}})

x \to \infty lim (sec (x))

f (x) = 5 x, at x = 4, x = 9

x \to - 3 - lim (\frac{3 - x}{9 - x ^{2}})