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

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

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integral of 1/(x^2-2)

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

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

Solution steps

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

Apply Integral Substitution

= \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

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

Factor

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

= \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

= \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}

= \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

x \to 3 lim (\frac{1}{1 + ( \frac{1}{x - 3} )})

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

y^{^{''}} + 4 y = 3 x - 1

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

y^{^{'}} = - x

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