Given that
g(x)=4x2−13x+13x−3 g(x) = \frac{4x^2 - 13x + 13}{x - 3} g(x)=x−34x2−13x+13Write g(x)g(x)g(x) in the form
Ax+B+Cx−3 Ax + B + \frac{C}{x - 3} Ax+B+x−3Cwhere AAA, BBB, and CCC are integers to be found.
Hence use algebraic integration to show that
∫46g(x) dx=α+βln3 \int_{4}^{6} g(x) \, dx = \alpha + \beta \ln 3 ∫46g(x)dx=α+βln3where α\alphaα and betabetabeta are integers to be found.