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

The integral of (4-x)/(x^4-x^2) is 4/x+3/2 ln|x-1|+ln|x|-5/2 ln|x+1|+C

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

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

\frac{4}{x} + \frac{3}{2} ln ∣ x - 1 ∣ + ln ∣ x ∣ - \frac{5}{2} ln ∣ x + 1 ∣ + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

\frac{4}{x} - 2 ln ∣ x + 1 ∣ + 2 ln ∣ x - 1 ∣ - 1 + \frac{1}{2} x ln ∣ x + 1 ∣ - \frac{1}{2} x ln ∣ x - 1 ∣ - (\frac{1}{2} x ln ∣ x + 1 ∣ - \frac{1}{2} x ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ - ln ∣ x ∣ - 1) : \frac{4}{x} + \frac{3}{2} ln ∣ x - 1 ∣ + ln ∣ x ∣ - \frac{5}{2} ln ∣ x + 1 ∣

= \frac{4}{x} + \frac{3}{2} ln ∣ x - 1 ∣ + ln ∣ x ∣ - \frac{5}{2} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{4}{x} + \frac{3}{2} ln ∣ x - 1 ∣ + ln ∣ x ∣ - \frac{5}{2} ln ∣ x + 1 ∣ + C

x \to 2 lim (2 x + c)

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

\frac{d y}{d x} = ∣ x - 2 ∣

f (x) = (x^{2} - 4)^{2}

\frac{d}{d t} (a cos (t))

What is the integral of (4-x)/(x^4-x^2) ?
The integral of (4-x)/(x^4-x^2) is 4/x+3/2 ln|x-1|+ln|x|-5/2 ln|x+1|+C