What is the integral of 1/(3x+12) ?

The integral of 1/(3x+12) is 1/3 ln|3x+12|+C

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integral of 1/(3x+12)

\int \frac{1}{3 x + 12} d x

\frac{1}{3} ln ∣ 3 x + 12 ∣ + C

Solution steps

\int \frac{1}{3 x + 12} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x + 12

= \frac{1}{3} ln ∣ 3 x + 12 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 x + 12 ∣ + C

\frac{\partial}{\partial y} (e^{x + y + 4})

\frac{d y}{d x} = 3 y + 1

\frac{d}{d x} (2 + cos (2 x))

x sin (x)

y = (1 + 4 x)^{\frac{3}{2}}

What is the integral of 1/(3x+12) ?
The integral of 1/(3x+12) is 1/3 ln|3x+12|+C