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

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

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integral of 1/(3x+2)

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

\frac{1}{3} ln ∣ 3 x + 2 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x + 2

= \frac{1}{3} ln ∣ 3 x + 2 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 x + 2 ∣ + C

\int_{0}^{2} (x - 3)^{2} - (x - 1)^{2} d x

(x + 1) y^{^{'}} + x y = 0

\frac{d}{d x} (\frac{2 x ^{2} - 1}{3 x ^{3} - 1})

\int e^{- x} sin^{2} (x) d x

\frac{d x}{d t} = \frac{4 x}{t}, x (1) = 1