What is the integral of (3x+18)/(x^2+8x+12) ?

The integral of (3x+18)/(x^2+8x+12) is 3ln|x+2|+C

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integral of (3x+18)/(x^2+8x+12)

\int \frac{3 x + 18}{x ^{2} + 8 x + 12} d x

3 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{3 x + 18}{x ^{2} + 8 x + 12} d x

Take the partial fraction of

\frac{3 x + 18}{x ^{2} + 8 x + 12} : \frac{3}{x + 2}

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{x + 2} d x

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 ln ∣ u ∣

Substitute back

u = x + 2

= 3 ln ∣ x + 2 ∣

Add a constant to the solution

= 3 ln ∣ x + 2 ∣ + C

What is the integral of (3x+18)/(x^2+8x+12) ?
The integral of (3x+18)/(x^2+8x+12) is 3ln|x+2|+C