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

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

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