What is the integral of 2/(1-4x^2) ?

The integral of 2/(1-4x^2) is 1/2 ln|2x+1|-1/2 ln|2x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/(1-4x^2)

\int \frac{2}{1 - 4 x ^{2}} d x

\frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{2}{1 - 4 x ^{2}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{1 - 4 x ^{2}} d x

Apply Integral Substitution

= 2 \cdot \int \frac{1}{2 ( 1 - u ^{2} )} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= 2 \cdot \frac{1}{2} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = 2 x

= 2 \cdot \frac{1}{2} (\frac{ln ∣ 2 x + 1 ∣}{2} - \frac{ln ∣ 2 x - 1 ∣}{2})

Simplify

2 \cdot \frac{1}{2} (\frac{ln ∣ 2 x + 1 ∣}{2} - \frac{ln ∣ 2 x - 1 ∣}{2}) : \frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2} ln ∣ 2 x - 1 ∣

= \frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x + 1 ∣ - \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

f (x) = 2 x - x^{2}, at x = 1

\frac{d}{d x} (\frac{x ^{3} - 1}{ln ( x )})

\int cos (x) sin (x) 1 + cos^{2} (x) d x

4 y^{^{''}} - 16 y = 0

\int 4 tan (x) d x

What is the integral of 2/(1-4x^2) ?
The integral of 2/(1-4x^2) is 1/2 ln|2x+1|-1/2 ln|2x-1|+C