What is the integral of 1/(4x-2) ?

The integral of 1/(4x-2) is 1/4 ln|4x-2|+C

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integral of 1/(4x-2)

\int \frac{1}{4 x - 2} d x

\frac{1}{4} ln ∣ 4 x - 2 ∣ + C

Solution steps

\int \frac{1}{4 x - 2} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 4 x - 2

= \frac{1}{4} ln ∣ 4 x - 2 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 x - 2 ∣ + C

x \to 3 lim (\frac{2}{x} + 1)

y = 5 x, y = \frac{5}{x}, x = 5

\int_{0}^{1} \frac{x}{e ^{6 x}} d x

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

y^{2} d x - 3 x d y = 0