What is the integral of (6x+9)/(3x^2+9x) ?

The integral of (6x+9)/(3x^2+9x) is 2ln|x+3|-ln| 2/3 x+2|+ln(2/3 |x|)+C

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integral of (6x+9)/(3x^2+9x)

\int \frac{6 x + 9}{3 x ^{2} + 9 x} d x

2 ln ∣ x + 3 ∣ - ln \frac{2}{3} x + 2 + ln (\frac{2}{3} ∣ x ∣) + C

Solution steps

\int \frac{6 x + 9}{3 x ^{2} + 9 x} d x

Expand

\frac{6 x + 9}{3 x ^{2} + 9 x} : \frac{6 x}{3 x ^{2} + 9 x} + \frac{9}{3 x ^{2} + 9 x}

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{6 x}{3 x ^{2} + 9 x} d x + \int \frac{9}{3 x ^{2} + 9 x} d x

\int \frac{6 x}{3 x ^{2} + 9 x} d x = 2 ln ∣ x + 3 ∣

\int \frac{9}{3 x ^{2} + 9 x} d x = - ln \frac{1}{3} (2 x + 3) + 1 + ln \frac{1}{3} (2 x + 3) - 1

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

Simplify

2 ln ∣ x + 3 ∣ - ln \frac{1}{3} (2 x + 3) + 1 + ln \frac{1}{3} (2 x + 3) - 1 : 2 ln ∣ x + 3 ∣ - ln \frac{2}{3} x + 2 + ln (\frac{2}{3} ∣ x ∣)

= 2 ln ∣ x + 3 ∣ - ln \frac{2}{3} x + 2 + ln (\frac{2}{3} ∣ x ∣)

Add a constant to the solution

= 2 ln ∣ x + 3 ∣ - ln \frac{2}{3} x + 2 + ln (\frac{2}{3} ∣ x ∣) + C

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What is the integral of (6x+9)/(3x^2+9x) ?
The integral of (6x+9)/(3x^2+9x) is 2ln|x+3|-ln| 2/3 x+2|+ln(2/3 |x|)+C