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

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

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

\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 + C

Solution steps

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

Apply Integral Substitution

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

Factor

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

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

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

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

Apply the Sum Rule:

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

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

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

\int 1 d u = u

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

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