What is the integral of 6/(1+6x) ?

The integral of 6/(1+6x) is ln|1+6x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6/(1+6x)

\int \frac{6}{1 + 6 x} d x

ln ∣ 1 + 6 x ∣ + C

Solution steps

\int \frac{6}{1 + 6 x} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{1 + 6 x} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= 6 \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 1 + 6 x

= 6 \cdot \frac{1}{6} ln ∣ 1 + 6 x ∣

Simplify

6 \cdot \frac{1}{6} ln ∣ 1 + 6 x ∣ : ln ∣ 1 + 6 x ∣

= ln ∣ 1 + 6 x ∣

Add a constant to the solution

= ln ∣ 1 + 6 x ∣ + C

\int 81 - t^{2} d t

\int \frac{sin ( 2 x )}{cos ( 2 x )} d x

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

x \to \infty lim (- x^{2})

\frac{d}{d t} (1 + sin (t))