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

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

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

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

Solution steps

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

Apply Integral Substitution

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} 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 = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{( ( x - 4 ) ^{7} )}{e ^{x}}, 4

\frac{1}{7 x}

\frac{\partial}{\partial x} (sin (8 π t + 9 x))

\frac{\partial}{\partial x} (x^{2} + 10 x yz + y^{2} + 8 z^{2})

\frac{d}{d x} (arctanh (cos (x)))

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