Question

Calcule o valor da integral indefinida usando o método para decompor em frações parciais.
teorema a ser usado: ∫$\frac{P(x)}{(x-\alpha)(x-\beta ) }$= A In $|x-\alpha |+BIn|x-\beta |+k$

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm integrac_{\!\!,}\tilde ao~de~func_{\!\!,}\tilde oes~racionais}\\\underline{\rm por~frac_{\!\!,}\tilde oes~parciais}\\\sf\dfrac{x}{x^2-5x+6}=\dfrac{x}{(x-2)(x-3)}=\dfrac{A}{x-2}+\dfrac{B}{x-3}\\\\\sf A=\dfrac{x}{x-3}\bigg|_{x=2}=\dfrac{2}{2-3}=-2\\\\\sf B=\dfrac{x}{x-2}\bigg |_{x=3}=\dfrac{3}{3-2}=3\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{x}{x^2-5x+6}~dx=-2\ell n|x-2|+3\ell n|x-3|+c\\\displaystyle\sf\int\dfrac{x}{x^2-5x+6}~dx=\ell n\bigg|\dfrac{(x-3)^3}{(x-2)^2}\bigg|+c\end{array}}$

Answer 2

Explicação passo-a-passo:

$\Large \: \int \frac{x}{ {x}^{2} - 5x + 6 } dx \\ \\ {x}^{2} - 5x + 6 = 0 \\ \\ x1 + x2 = \frac{ - b}{a} = \frac{ - ( - 5)}{1} = 5 \\ \\ x1 \times x2 = \frac{c}{a} = \frac{6}{1} = 6 \\ x1 = 2 \\ x2 = 3 \\ \\ a( x - x1)(x - x2) \\ \\ 1(x - 2)(x - 3) = (x - 2)(x - 3)\\ \\ \frac{x}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3} \\ \\ \frac{x = (x - 3)A + (x - 2)B}{(x - 2)(x - 3)} \\ \\ x = Ax - 3A + Bx - 2B \\ \\ x = Ax + Bx - 3A- 2B \\ \\ x = (A+ B)x + ( - 3A- 2B) \\ \\ \left \{ {({A+ B = 1} ) \times 2\atop{- 3A- 2B = 0}}\right. \\ \\ \left \{ {{2A+ 2B = 2} \atop{- 3A- 2B = 0}}\right. \\ - - - - - - - - - \boxed{ + } \\ - A = 2 \\ \\ \Large \boxed{A = - 2} \\ \\ A+B=1 \\ \\ - 2 + B = 1 \\ \\ B = 1 + 2 \\ \\ \Large \boxed{B = 3} \\ \\ \Large \int \frac{ - 2}{x - 2} + \frac{3}{x - 3} dx \\ \\ \Large \int \frac{ - 2}{x - 2} dx + \Large \int\frac{3}{x - 3} dx \\ \\ \Large \boxed{ \green{- 2ln |x - 2| + 3ln |x - 3| + k,\: k \in \mathbb{R} }}$

$\Large \boxed{\underline{\bf \blue{Bons\: Estudos!}}}\\ \\\Large \boxed{\underline{\bf 28/05/2021}}$