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

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

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

ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

Solution steps

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

Use the common integral:

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

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

Simplify

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

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣ + C

\int - sin^{2} (t) d t

y^{^{'}} + sin (t) y + sin (t) = 0, y (0) = 6

y^{^{''}} + 25 y = 40 sin (5 x)

2 (x + 6), - 6, - 2

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