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

The integral of (x^2+1)/(x^2-2) is x-3/(2sqrt(2))ln| x/(sqrt(2))+1|+3/(2sqrt(2))ln| x/(sqrt(2))-1|+C

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integral of (x^2+1)/(x^2-2)

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

x - \frac{3}{2 2} ln \frac{x}{2} + 1 + \frac{3}{2 2} ln \frac{x}{2} - 1 + C

Solution steps

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

Expand

\frac{x ^{2} + 1}{x ^{2} - 2} : \frac{x ^{2}}{x ^{2} - 2} + \frac{1}{x ^{2} - 2}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

- \frac{1}{2} ln \frac{x}{2} + 1 + \frac{1}{2} ln \frac{x}{2} - 1 + x - \frac{1}{2 2} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) : x - \frac{3}{2 2} ln \frac{x}{2} + 1 + \frac{3}{2 2} ln \frac{x}{2} - 1

= x - \frac{3}{2 2} ln \frac{x}{2} + 1 + \frac{3}{2 2} ln \frac{x}{2} - 1

Add a constant to the solution

= x - \frac{3}{2 2} ln \frac{x}{2} + 1 + \frac{3}{2 2} ln \frac{x}{2} - 1 + C

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What is the integral of (x^2+1)/(x^2-2) ?
The integral of (x^2+1)/(x^2-2) is x-3/(2sqrt(2))ln| x/(sqrt(2))+1|+3/(2sqrt(2))ln| x/(sqrt(2))-1|+C