$$\begin{align}& \end{align}$$
¡Hola Alejandro!
·
$$\begin{align}&\int \frac {8x^2+4x-2}{x^3+x^2 } dx=\\&\\&\text{vamos a calcular las fracciones parciales}\\&\\&\frac {8x^2+4x-2}{x^3+x^2 }=\frac {8x^2+4x-2}{x^2(x+1)}=\\&\\&\text{las fracciones correspondientes son}\\&\\&=\frac ax+ \frac b{x^2}+\frac c{x+1} =\\&\\&\frac{ax(x+1)+b(x+1)+cx^2}{x^2(x+1)}\\&\\&\text{ese numerado es igual al original}\\&\\&8x^2+4x-2 = ax(x+1)+b(x+1)+cx^2\\&\text{tomando x=0}\\&-2=b\implies b=-2\\&\text{tomando x=-1}\\&8-4-2=c\implies c=2\\&\text{tomando x=1}\\&8+4-2=2a+2b+c =2a-4+2\\&10=2a-2\implies2a=12\implies a=6\\&\\&\text{y retomamos la integral}\\&\\&=6\int \frac{dx}{x}-2\int \frac{dx}{x^2}+2\int \frac{dx}{x+1}=\\&\\&6\,ln|x|+ \frac 2x+2\, ln|x+1|+x+C\end{align}$$
Y eso es todo.