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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{3}{2 ( x + 1 )} d x = \frac{3}{2} ln ∣ x + 1 ∣

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

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

Add a constant to the solution

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

x \to 5 - lim (\frac{1}{5} x + 4)

\frac{d}{d x} ((x^{2} - 3)^{4})

f (x) = x + 8

(e^{2 x})^{^{'}}

e^{- 6 t}

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