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

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

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

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

3 (x + 6 - 6 ln ∣ x + 6 ∣) + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{x}{x + 6} d x

Apply u-substitution

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

= 3 \cdot \int \frac{u - 6}{u} d u

Expand

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

= 3 \cdot \int 1 - \frac{6}{u} d u

Apply the Sum Rule:

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

= 3 (\int 1 d u - \int \frac{6}{u} d u)

\int 1 d u = u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

= 3 (u - 6 ln ∣ u ∣)

Substitute back

u = x + 6

= 3 (x + 6 - 6 ln ∣ x + 6 ∣)

Add a constant to the solution

= 3 (x + 6 - 6 ln ∣ x + 6 ∣) + C

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