Question

`\rm \int_0^1 \frac{ {x}^(3) - {x}^(2) - 1 }{ {x}^(2) - x - 1 }dx \\`​

Answer 1

Simplify the integrand.

`\displaystyle (x^3-x^2-1)/(x^2-x-1) = x + (x-1)/(x^2-x-1)`

Expand into partial fractions, recalling that
`\phi` and
`-\frac1\phi` are roots to the quadratic in the denominator.

`(x-1)/((x - \phi) \left(x + \frac1\phi\right)) = \frac1{1+\phi^2} \left(\frac1{x-\phi} + (\phi^2)/(x + \frac1\phi)\right)`

Then the integral is

`\displaystyle I = \int_0^1 (x^3-x^2-1)/(x^2-x-1) \, dx \\\\ ~~~~ = \int_0^1 x \, dx + \frac1{1+\phi^2} \int_0^1 \left(\frac1{x-\phi} + (\phi^2)/(x+\frac1\phi)\right) \, dx \\\\ ~~~~ = \frac12 + (1)/(1+\phi^2) \left(\ln|1-\phi| + \phi^2 \ln\left|1 + \frac1\phi\right| - \ln|-\phi| - \phi^2 \ln\left|\frac1\phi\right|\right)`

and this reduces significantly, when we recall that
`\phi = 1 + \frac1\phi`, to

`\displaystyle I = \boxed{\frac12 + (2\phi)/(1 + \phi^2) \ln(\phi)}`