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

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

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

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

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

Solution steps

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

Expand

\frac{x - 1}{x ^{2} - x - 1} : \frac{x}{x ^{2} - x - 1} - \frac{1}{x ^{2} - x - 1}

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

\int \frac{x}{x ^{2} - x - 1} d x = - 2 (- \frac{1}{4} ln - 4 x^{2} + 4 x + 4 + \frac{1}{4 5} (ln \frac{2 x - 1}{5} + 1 - ln \frac{2 x - 1}{5} - 1))

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

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

Simplify

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

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

Add a constant to the solution

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

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