What is the integral of (2x^3-x^2+4x-4)/(x^4-4x^2) ?

The integral of (2x^3-x^2+4x-4)/(x^4-4x^2) is

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integral of (2x^3-x^2+4x-4)/(x^4-4x^2)

\int \frac{2 x ^{3} - x ^{2} + 4 x - 4}{x ^{4} - 4 x ^{2}} d x

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

Solution steps

\int \frac{2 x ^{3} - x ^{2} + 4 x - 4}{x ^{4} - 4 x ^{2}} d x

Expand

\frac{2 x ^{3} - x ^{2} + 4 x - 4}{x ^{4} - 4 x ^{2}} : \frac{2 x ^{3}}{x ^{4} - 4 x ^{2}} - \frac{x ^{2}}{x ^{4} - 4 x ^{2}} + \frac{4 x}{x ^{4} - 4 x ^{2}} - \frac{4}{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{2 x ^{3}}{x ^{4} - 4 x ^{2}} d x - \int \frac{x ^{2}}{x ^{4} - 4 x ^{2}} d x + \int \frac{4 x}{x ^{4} - 4 x ^{2}} d x - \int \frac{4}{x ^{4} - 4 x ^{2}} d x

\int \frac{2 x ^{3}}{x ^{4} - 4 x ^{2}} d x = ln x^{2} - 4

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{2} 1 0^{x} d x

\frac{d}{d x} (e^{3 x} - 1 - 3 x)

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

\frac{d}{d x} (x^{5} - 20 x^{3} + 64 x)

What is the integral of (2x^3-x^2+4x-4)/(x^4-4x^2) ?
The integral of (2x^3-x^2+4x-4)/(x^4-4x^2) is