What is the integral of 4/((4x-1)ln(4x-1)) ?

The integral of 4/((4x-1)ln(4x-1)) is ln|ln(4x-1)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4/((4x-1)ln(4x-1))

\int \frac{4}{( 4 x - 1 ) ln ( 4 x - 1 )} d x

ln ∣ ln (4 x - 1) ∣ + C

Solution steps

\int \frac{4}{( 4 x - 1 ) ln ( 4 x - 1 )} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{( 4 x - 1 ) ln ( 4 x - 1 )} d x

Apply u-substitution

: \int \frac{1}{4 u ln ( u )} d u

= 4 \cdot \int \frac{1}{4 u ln ( u )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{u ln ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= 4 \cdot \frac{1}{4} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 4 \cdot \frac{1}{4} ln ∣ v ∣

Substitute back

= 4 \cdot \frac{1}{4} ln ∣ ln (4 x - 1) ∣

Simplify

4 \cdot \frac{1}{4} ln ∣ ln (4 x - 1) ∣ : ln ∣ ln (4 x - 1) ∣

= ln ∣ ln (4 x - 1) ∣

Add a constant to the solution

= ln ∣ ln (4 x - 1) ∣ + C

\int \frac{1}{x ^{3}} (1 + \frac{1}{x ^{2}})^{\frac{5}{3}} d x

\int (8 x^{3} - 6 x^{- 2}) d x

\int \frac{3 x ^{2}}{x ^{3} + 5} d x

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

\int (x + 2)^{2} d x

What is the integral of 4/((4x-1)ln(4x-1)) ?
The integral of 4/((4x-1)ln(4x-1)) is ln|ln(4x-1)|+C