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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3/(3x+1)

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

ln ∣ 3 x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{3 x + 1} d x

Apply u-substitution

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

= 3 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x + 1

= 3 \cdot \frac{1}{3} ln ∣ 3 x + 1 ∣

Simplify

3 \cdot \frac{1}{3} ln ∣ 3 x + 1 ∣ : ln ∣ 3 x + 1 ∣

= ln ∣ 3 x + 1 ∣

Add a constant to the solution

= ln ∣ 3 x + 1 ∣ + C

\int \frac{x ^{2} + 57}{x ^{2} + 64} d x

\int 3 sin^{5} (2 x) d x

\int (2^{- 3 x^{2}}) x d x

\int \frac{1}{( 1 + tan ( x ) )} d x

\int \frac{x + x ^{3}}{1 + x ^{4}} d x