What is the integral of 1/(3-4x) ?

The integral of 1/(3-4x) is -1/4 ln|3-4x|+C

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integral of 1/(3-4x)

\int \frac{1}{3 - 4 x} d x

- \frac{1}{4} ln ∣ 3 - 4 x ∣ + C

Solution steps

\int \frac{1}{3 - 4 x} 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 = 3 - 4 x

= - \frac{1}{4} ln ∣ 3 - 4 x ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ 3 - 4 x ∣ + C

\frac{d}{d x} (\frac{x}{9.8})

f (x) = (ln (7)) e^{x}

f (x) = \frac{1}{x ^{3}}, (- 2, - \frac{1}{8})

\int x e^{\frac{x}{3}} d x

\frac{d y}{d x}, sin (x + y) = 3 x - 2 y