What is the integral of (9x+6)/(3x+1) ?

The integral of (9x+6)/(3x+1) is 3x+1+ln|3x+1|+C

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integral of (9x+6)/(3x+1)

\int \frac{9 x + 6}{3 x + 1} d x

3 x + 1 + ln ∣ 3 x + 1 ∣ + C

Solution steps

\int \frac{9 x + 6}{3 x + 1} d x

Expand

\frac{9 x + 6}{3 x + 1} : \frac{9 x}{3 x + 1} + \frac{6}{3 x + 1}

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{9 x}{3 x + 1} d x + \int \frac{6}{3 x + 1} d x

\int \frac{9 x}{3 x + 1} d x = 3 x + 1 - ln ∣ 3 x + 1 ∣

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

= 3 x + 1 - ln ∣ 3 x + 1 ∣ + 2 ln ∣ 3 x + 1 ∣

Simplify

= 3 x + 1 + ln ∣ 3 x + 1 ∣

Add a constant to the solution

= 3 x + 1 + ln ∣ 3 x + 1 ∣ + C

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What is the integral of (9x+6)/(3x+1) ?
The integral of (9x+6)/(3x+1) is 3x+1+ln|3x+1|+C