What is the integral of (x-5)/(-2x+2) ?

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

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integral of (x-5)/(-2x+2)

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

\frac{x ^{2}}{- 2 x + 2} - \frac{1}{2} (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) + \frac{5}{2} ln ∣ - 2 x + 2 ∣ + C

Solution steps

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

Expand

\frac{x - 5}{- 2 x + 2} : \frac{x}{- 2 x + 2} - \frac{5}{- 2 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 + 2} d x - \int \frac{5}{- 2 x + 2} d x

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

\int \frac{5}{- 2 x + 2} d x = - \frac{5}{2} ln ∣ - 2 x + 2 ∣

= \frac{x ^{2}}{- 2 x + 2} - \frac{1}{2} (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - (- \frac{5}{2} ln ∣ - 2 x + 2 ∣)

Simplify

= \frac{x ^{2}}{- 2 x + 2} - \frac{1}{2} (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) + \frac{5}{2} ln ∣ - 2 x + 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{- 2 x + 2} - \frac{1}{2} (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) + \frac{5}{2} ln ∣ - 2 x + 2 ∣ + C

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