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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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