Question

Calcule a integral :
Imagem anexada abaixo.

Answer 1

$\displaystyle \sf \int\frac{x^3}{x^2-3x+2}dx$

Vamos manipular o numerador de tal forma que dê pra simplificar ao máximo:

$\displaystyle \sf \int \left[\frac{x^3-3x^2+2x+3x^2-2x}{x^2-3x+2}\right]dx \\\\\\ \int \left[\frac{x\cdot (x^2-3x+2)}{x^2-3x+2}+\frac{3x^2-2x}{x^2-3x+2}\right]dx \\\\\\ \int\left[x+\frac{3x^2-9x+6+7x-6}{x^2-3x+2}\right]dx \\\\\\ \int\left[x+\frac{3\cdot(x^2-3x+2)}{x^2-3x+2}+\frac{7x-6}{x^2-3x+2}\right]dx \\\\\\ \int \left[x+3+\frac{7x-6}{x^2-3x+2}\right]dx$

Agora vamos achar as frações parciais da ultima fração.

$\displaystyle \sf \frac{7x-6}{x^2-3x+2} = \frac{7x-6}{(x-1)\cdot (x-2)}\\\\\\ \frac{7x-6}{(x-1)\cdot (x-2)}=\frac{A}{(x-1)}+\frac{B}{(x-2)} \\\\\\ \frac{7x-6}{(x-1)\cdot (x-2) }=\frac{A\cdot (x-1)+B\cdot (x-1)}{(x-1)\cdot (x-2) } \\\\\\ \frac{7x-6}{(x-1)\cdot(x-2)}=\frac{Ax+Bx-2A-B}{(x-1)\cdot(x-2)}$

$\displaystyle \left\{ \begin{array}{I} \sf (A+B)x=7x \to A+B = 7 \to B = 7-A \\\\ \sf -2A-B=-6 \to 2A+B=6 \to 2A+7-A =6 \end{array} \right \\\\\\ \left\{\begin{array}{I} \boxed{\sf A = -1}\\\\ \boxed{\sf B = 8} \end{array} \right$

Voltando para integral :

$\displaystyle \sf \int\left[x+3+\frac{7x-6}{x^2-3x+2} \right]dx \\\\\\ \int \left[x+3-\frac{1}{(x-1)}+\frac{8}{(x-2)} \right]dx \\\\\\ \int xdx + \int 3dx - \int \frac{1}{(x-1)} dx+8\int \frac{1}{(x-2)}dx \\\\\\ \boxed{\sf \frac{x^2}{2}+3x-\ln |x-1|+8 \ln |x-2|+C }\checkmark$