What is the integral of 1/(4-6x) ?

The integral of 1/(4-6x) is -1/6 ln|4-6x|+C

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integral of 1/(4-6x)

\int \frac{1}{4 - 6 x} d x

- \frac{1}{6} ln ∣ 4 - 6 x ∣ + C

Solution steps

\int \frac{1}{4 - 6 x} d x

Apply u-substitution

: \int - \frac{1}{6 u} d u

= \int - \frac{1}{6 u} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{6} ln ∣ u ∣

Substitute back

u = 4 - 6 x

= - \frac{1}{6} ln ∣ 4 - 6 x ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ 4 - 6 x ∣ + C

d x + d y + x d y = y d x

f (x) = \frac{2 π - x cos ( x )}{x - 2 π}

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

x \frac{d y}{d x} = 8 y

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