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

The integral of (3x-2)/(x+1) is 3(x+1)-5ln|x+1|+C

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

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

3 (x + 1) - 5 ln ∣ x + 1 ∣ + C

Solution steps

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

Apply u-substitution

= \int \frac{3 u - 5}{u} d u

Expand

\frac{3 u - 5}{u} : 3 - \frac{5}{u}

= \int 3 - \frac{5}{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{5}{u} d u

\int 3 d u = 3 u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= 3 u - 5 ln ∣ u ∣

Substitute back

u = x + 1

= 3 (x + 1) - 5 ln ∣ x + 1 ∣

Add a constant to the solution

= 3 (x + 1) - 5 ln ∣ x + 1 ∣ + C

x \to - 3 lim (∣ - x - 1 ∣)

\int \frac{1}{( x - 3 ) ( x - 1 )} d x

d er i v a t i v e \frac{6}{5} - \frac{6}{5} e^{- 20 t}

l o n g d i v i s i o n \frac{x ^{2} + 1}{x ^{2} - 4}

\frac{d y}{d x} = 4 - y

What is the integral of (3x-2)/(x+1) ?
The integral of (3x-2)/(x+1) is 3(x+1)-5ln|x+1|+C