What is the integral of (3x-1)/(x+2) ?

The integral of (3x-1)/(x+2) is 3(x+2)-7ln|x+2|+C

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integral of (3x-1)/(x+2)

\int \frac{3 x - 1}{x + 2} d x

3 (x + 2) - 7 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{3 x - 1}{x + 2} d x

Apply u-substitution

= \int \frac{3 u - 7}{u} d u

Expand

\frac{3 u - 7}{u} : 3 - \frac{7}{u}

= \int 3 - \frac{7}{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

= \int 3 d u - \int \frac{7}{u} d u

\int 3 d u = 3 u

\int \frac{7}{u} d u = 7 ln ∣ u ∣

= 3 u - 7 ln ∣ u ∣

Substitute back

u = x + 2

= 3 (x + 2) - 7 ln ∣ x + 2 ∣

Add a constant to the solution

= 3 (x + 2) - 7 ln ∣ x + 2 ∣ + C

What is the integral of (3x-1)/(x+2) ?
The integral of (3x-1)/(x+2) is 3(x+2)-7ln|x+2|+C