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

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

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integral of 1/(x^2-n)

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

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

Solution steps

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

Apply Integral Substitution

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

Factor

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

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

Substitute back

u = \frac{x}{n}

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

Simplify

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

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

Add a constant to the solution

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

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