Question

Usando o método de decomposição em frações parciais, integre a função:

Answer 1

(1) Descomposición en fracciones parciales

$\dfrac{3x+1}{x^2-5x+6}=\dfrac{3x+1}{(x-2)(x-3)}\\ \\ \\ \dfrac{3x+1}{x^2-5x+6}=\dfrac{A}{x-2}+\dfrac{B}{x-3}\\ \\ \\ \dfrac{3x+1}{x^2-5x+6}=\dfrac{(A+B)x-3A-2B}{(x-2)(x-3)}\\ \\ \\ 3x+1=(A+B)x-3A-2B\\ \\ \\ \begin{cases} A+B=3\\ 3A+2B=-1 \end{cases}\\ \\ \\ \boxed{A=-7~~~\wedge~~~ B=10}$

                    $\dfrac{3x+1}{x^2-5x+6}=-\dfrac{7}{x-2}+\dfrac{10}{x-3}$

(2) Integración

$\displaystyle \int\dfrac{3x+1}{x^2-5x+6}dx=\int-\dfrac{7}{x-2}+\dfrac{10}{x-3}\;dx\\ \\ \\ \int\dfrac{3x+1}{x^2-5x+6}dx=-7\int\dfrac{dx}{x-2}+10\int\dfrac{dx}{x-3}\\ \\ \\ \int\dfrac{3x+1}{x^2-5x+6}dx=-7\ln|x-2|+10\ln|x-3|+C\\ \\ \\ ~~~~~~~~~~~~~~~~\boxed{\int\dfrac{3x+1}{x^2-5x+6}dx=\ln\left|\dfrac{(x-3)^{10}}{(x-2)^7}\right|+C}$