What is the integral of 8/(1+8x) ?

The integral of 8/(1+8x) is ln|1+8x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/(1+8x)

\int \frac{8}{1 + 8 x} d x

ln ∣ 1 + 8 x ∣ + C

Solution steps

\int \frac{8}{1 + 8 x} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{1 + 8 x} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= 8 \cdot \int \frac{1}{8 u} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{8} ln ∣ u ∣

Substitute back

u = 1 + 8 x

= 8 \cdot \frac{1}{8} ln ∣ 1 + 8 x ∣

Simplify

8 \cdot \frac{1}{8} ln ∣ 1 + 8 x ∣ : ln ∣ 1 + 8 x ∣

= ln ∣ 1 + 8 x ∣

Add a constant to the solution

= ln ∣ 1 + 8 x ∣ + C

\frac{d}{d x} ((3 x^{2})^{2 x} \cdot (x^{2} + 2 x)^{2})

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

\int tan^{3} (θ) sec^{4} (θ) d θ

\int (4 - x^{3} - 5 x^{5}) d x

x \to 0 + lim (tan^{x} (6 x))