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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2-x^2)

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

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

Solution steps

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

Apply Integral Substitution

= \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

= \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}

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

n = 4 \sum \infty \frac{8}{n ^{2} - 1}

2 x^{2}, 2 x + 6, - 1, 0

x y + 4 y = 10

x \to 2 lim (\frac{5 x ^{3}}{x + 3})

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

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