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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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