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

The integral of (x+1)/(3x+2) is 1/3 x+2/9+1/9 ln|3x+2|+C

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

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

\frac{1}{3} x + \frac{2}{9} + \frac{1}{9} ln ∣ 3 x + 2 ∣ + C

Solution steps

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

Expand

\frac{x + 1}{3 x + 2} : \frac{x}{3 x + 2} + \frac{1}{3 x + 2}

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

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

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

= \frac{1}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣) + \frac{1}{3} ln ∣ 3 x + 2 ∣

Simplify

\frac{1}{9} (3 x + 2 - 2 ln ∣ 3 x + 2 ∣) + \frac{1}{3} ln ∣ 3 x + 2 ∣ : \frac{1}{3} x + \frac{2}{9} + \frac{1}{9} ln ∣ 3 x + 2 ∣

= \frac{1}{3} x + \frac{2}{9} + \frac{1}{9} ln ∣ 3 x + 2 ∣

Add a constant to the solution

= \frac{1}{3} x + \frac{2}{9} + \frac{1}{9} ln ∣ 3 x + 2 ∣ + C

\int_{- 3}^{1} x d x

x \to \infty lim (2 ln (x))

x \to - 6 lim (5 x^{2} + 6 x + a)

\int_{0}^{e} \frac{1}{x + e} d x

\int \frac{x}{1 + 9 x ^{2}} d x

What is the integral of (x+1)/(3x+2) ?
The integral of (x+1)/(3x+2) is 1/3 x+2/9+1/9 ln|3x+2|+C