What is the integral of (x^2)/(x^4-2x^2-8) ?

The integral of (x^2)/(x^4-2x^2-8) is 1/(3sqrt(2))arctan(x/(sqrt(2)))-1/6 ln|x+2|+1/6 ln|x-2|+C

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integral of (x^2)/(x^4-2x^2-8)

\int \frac{x ^{2}}{x ^{4} - 2 x ^{2} - 8} d x

\frac{1}{3 2} arctan (\frac{x}{2}) - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{2}}{x ^{4} - 2 x ^{2} - 8} d x

Take the partial fraction of

\frac{x ^{2}}{x ^{4} - 2 x ^{2} - 8} : \frac{1}{3 ( x ^{2} + 2 )} - \frac{1}{6 ( x + 2 )} + \frac{1}{6 ( x - 2 )}

= \int \frac{1}{3 ( x ^{2} + 2 )} - \frac{1}{6 ( x + 2 )} + \frac{1}{6 ( x - 2 )} d x

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{1}{3 ( x ^{2} + 2 )} d x - \int \frac{1}{6 ( x + 2 )} d x + \int \frac{1}{6 ( x - 2 )} d x

\int \frac{1}{3 ( x ^{2} + 2 )} d x = \frac{1}{3 2} arctan (\frac{x}{2})

\int \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} ln ∣ x + 2 ∣

\int \frac{1}{6 ( x - 2 )} d x = \frac{1}{6} ln ∣ x - 2 ∣

= \frac{1}{3 2} arctan (\frac{x}{2}) - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{1}{3 2} arctan (\frac{x}{2}) - \frac{1}{6} ln ∣ x + 2 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + C

\frac{\partial}{\partial x} (π x^{2} y)

n \to \infty lim (\frac{1}{6})

\int (3 cos (3 x) - 5 sin (5 x)) d x

f (x) = 3 - x

\int_{9}^{1} (- 15 x) d x

What is the integral of (x^2)/(x^4-2x^2-8) ?
The integral of (x^2)/(x^4-2x^2-8) is 1/(3sqrt(2))arctan(x/(sqrt(2)))-1/6 ln|x+2|+1/6 ln|x-2|+C