What is the integral of 1/(7x+3) ?

The integral of 1/(7x+3) is 1/7 ln|7x+3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(7x+3)

\int \frac{1}{7 x + 3} d x

\frac{1}{7} ln ∣ 7 x + 3 ∣ + C

Solution steps

\int \frac{1}{7 x + 3} d x

Apply u-substitution

: \int \frac{1}{7 u} d u

= \int \frac{1}{7 u} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{7} ln ∣ u ∣

Substitute back

u = 7 x + 3

= \frac{1}{7} ln ∣ 7 x + 3 ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ 7 x + 3 ∣ + C

\frac{d}{d t} (\frac{2 t + 7}{2 t})

\frac{d}{d x} (- 2 tan (x))

\int (4 x - 1)^{3} d x

\int_{1}^{6} (2 x^{3} + 5 x^{2} + 2 x + 8) d x

x \to 3 + lim (\frac{1}{6 - 2 x})