Question

Usando o método de decomposição em frações parciais, integre a função: ∫ 3x + 1 / x² - 5x + 6 dx

Answer 1

 Como sugerido pelo enunciado,

$\\ \mathsf{\frac{3x + 1}{x^2 - 5x + 6} = \frac{A}{x - 2} + \frac{B}{x - 3}} \\\\\\ \mathsf{\frac{3x + 1}{x^2 - 5x + 6} = \frac{A(x - 3) + B(x - 2)}{(x - 2)(x - 3)}} \\\\\\ \mathsf{\frac{3x + 1}{x^2 - 5x + 6} = \frac{Ax - 3A + Bx - 2B}{(x - 2)(x - 3)}} \\\\\\ \mathsf{\frac{3x + 1}{x^2 - 5x + 6} = \frac{(A + B)x + (- 3A - 2B)}{(x - 2)(x - 3)}}$
 
 Por comparação,

$\\ \begin{cases} \mathsf{A + B = 3 \qquad \qquad \times(2} \\ \mathsf{- 3A - 2B = 1}\end{cases} \\\\\\ \begin{cases} \mathsf{2A + 2B = 6} \\ \mathsf{- 3A - 2B = 1}\end{cases} \\ -------- \\ \mathsf{- A = 7} \\\\ \boxed{\mathsf{A = - 7}}$
 
 Substituindo,

$\\ \mathsf{A + B = 3} \\\\ \mathsf{- 7 + B = 3} \\\\ \boxed{\mathsf{B = 10}}$
 
 Desse modo, temos que:

$\\ \mathsf{\int \frac{3x + 1}{x^2 - 5x + 6} \ dx = \int \left (\frac{- 7}{x - 2} + \frac{10}{x - 3} \right ) \ dx} \\\\\\ \mathsf{\int \frac{3x + 1}{x^2 - 5x + 6} \ dx = \int \frac{- 7}{x - 2} \ dx + \int \frac{10}{x - 3} \ dx} \\\\\\ \mathsf{\int \frac{3x + 1}{x^2 - 5x + 6} \ dx = - 7 \cdot \int \frac{1}{x - 2} \ dx + 10 \cdot \int \frac{1}{x - 3} \ dx} \\\\\\ \boxed{\boxed{\mathsf{\int \frac{3x + 1}{x^2 - 5x + 6} \ dx = - 7 \cdot \ln |x - 2| + 10 \cdot \ln |x - 3| + C}}}$