What is the integral of (x^2+1)/(2x-x^2) ?

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

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integral of (x^2+1)/(2x-x^2)

\int \frac{x ^{2} + 1}{2 x - x ^{2}} d x

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

Solution steps

\int \frac{x ^{2} + 1}{2 x - x ^{2}} d x

Expand

\frac{x ^{2} + 1}{2 x - x ^{2}} : \frac{x ^{2}}{2 x - x ^{2}} + \frac{1}{2 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{x ^{2}}{2 x - x ^{2}} d x + \int \frac{1}{2 x - x ^{2}} d x

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

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

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

Simplify

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

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

Add a constant to the solution

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

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