integration by partial fractions

Question

integral ((6x^2+3x+6)/(x^3-1))

Answer 1

(6x² + 3x + 6)/(x - 1)(x² + x + 1) = a/(x - 1) + (bx + c)/(x² + x + 1); 6x² + 3x + 6 = a(x² + x + 1) + (bx + c)(x - 1);

a + b = 6; a - b + c = 3; a - c = 6; a = 5, b = 1, c = - 1

Integral I = ∫5/(x - 1)dx + ∫(x - 1)/(x² - x + 1)dx = 5lnIx - 1I + 1/2∫(2x - 1)/(x² - x + 1) - 1/2∫1/[(x - 1/2)² + 3/4]dx =

5lnIx - 1I + 1/2lnIx² - x + 1I - 1/√3tan^-1[(2x - 1)/√3] + C