What is the integral of (x-2)/(3x+6) ?

The integral of (x-2)/(3x+6) is 1/3 x-4/3 ln|3x+6|+C

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integral of (x-2)/(3x+6)

\int \frac{x - 2}{3 x + 6} d x

\frac{1}{3} x - \frac{4}{3} ln ∣ 3 x + 6 ∣ + C

Solution steps

\int \frac{x - 2}{3 x + 6} d x

Expand

\frac{x - 2}{3 x + 6} : \frac{x}{3 x + 6} - \frac{2}{3 x + 6}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{3 x + 6} d x - \int \frac{2}{3 x + 6} d x

\int \frac{x}{3 x + 6} d x = \frac{1}{3} x - \frac{2}{3} ln ∣ 3 x + 6 ∣

\int \frac{2}{3 x + 6} d x = \frac{2}{3} ln ∣ 3 x + 6 ∣

= \frac{1}{3} x - \frac{2}{3} ln ∣ 3 x + 6 ∣ - \frac{2}{3} ln ∣ 3 x + 6 ∣

Add similar elements:

- \frac{2}{3} ln ∣ 3 x + 6 ∣ - \frac{2}{3} ln ∣ 3 x + 6 ∣ = - \frac{4}{3} ln ∣ 3 x + 6 ∣

= \frac{1}{3} x - \frac{4}{3} ln ∣ 3 x + 6 ∣

Add a constant to the solution

= \frac{1}{3} x - \frac{4}{3} ln ∣ 3 x + 6 ∣ + C

\frac{d}{d x} (16 - x^{2} - y^{2})

\int \frac{x ^{3}}{x ^{2} - 36} d x

\frac{\partial}{\partial x} (e^{2 x} \cdot cos (x + y))

y = ∣ 2 x ∣, y = x^{2} - 3

\frac{d}{d x} (e^{- 2 x} (sin (3 x) + cos (3 x)))

What is the integral of (x-2)/(3x+6) ?
The integral of (x-2)/(3x+6) is 1/3 x-4/3 ln|3x+6|+C