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

The integral of (x^2+2x+8)/(x^4-4x^2) is

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

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

- \frac{1}{4} ln \frac{x}{2} + 1 + \frac{1}{4} ln \frac{x}{2} - 1 - \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + \frac{2}{x} + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{x ^{4} - 4 x ^{2}} d x = - \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

\int \frac{2 x}{x ^{4} - 4 x ^{2}} d x = 2 (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣)

\int \frac{8}{x ^{4} - 4 x ^{2}} d x = 8 (\frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣)

= - \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + 2 (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣) + 8 (\frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣)

Simplify

- \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + 2 (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{1}{8} ln ∣ x - 2 ∣) + 8 (\frac{1}{4 x} - \frac{1}{16} ln ∣ x + 2 ∣ + \frac{1}{16} ln ∣ x - 2 ∣) : - \frac{1}{4} ln \frac{x}{2} + 1 + \frac{1}{4} ln \frac{x}{2} - 1 - \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + \frac{2}{x}

= - \frac{1}{4} ln \frac{x}{2} + 1 + \frac{1}{4} ln \frac{x}{2} - 1 - \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + \frac{2}{x}

Add a constant to the solution

= - \frac{1}{4} ln \frac{x}{2} + 1 + \frac{1}{4} ln \frac{x}{2} - 1 - \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + \frac{2}{x} + C

\int_{1}^{2} \frac{- 1 ln ( x )}{x ^{2}} d x

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

n \to \infty lim (1.0 4^{n})

\frac{d}{d x} (- x^{3} + 6 x^{2} - 9 x)

\int \frac{- 5}{x ^{2} + 16} d x

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