Answer
$\displaystyle \frac{1}{2}+\ln 6$
Work Step by Step
$\displaystyle \frac{x^{3}+4x^{2}+x-1}{x^{3}+x^{2}}=\frac{(x^{3}+x^{2})+3x^{2}+x-1}{x^{3}+x^{2}}=1+\frac{3x^{2}+x-1}{x^{3}+x^{2}}$
$=1+\displaystyle \frac{3x^{2}+x-1}{x^{2}(x+1)}$
repeated factor
$=1 +\displaystyle \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$
$3x^{2}+x-1=Ax(x+1)+B(x+1)+Cx^{2}$
$3x^{2}+x-1=Ax^{2}+Ax+Bx+B+Cx^{2}$
$3x^{2}+x-1=(A+C)x^{2}+(A+B)x+B\Rightarrow\left[\begin{array}{lll}
B=-1 & & \\
A+B=1 & \Rightarrow A=2 & \\
A+C=3 & & \Rightarrow C=1
\end{array}\right]$
$\displaystyle \int_{1}^{2}\frac{x^{3}+4x^{2}+x-1}{x^{3}+x^{2}}dx=\int_{1}^{2}(1 +\frac{2}{x}-\frac{1}{x^{2}}+\frac{1}{x+1})dx$
$=\left[x+2\ln|x|+\dfrac{1}{x}+\ln|x+1|\right]_{1}^{2}$
$=(2+2\displaystyle \ln 2+\frac{1}{2}+\ln 3)-(1+0+1+\ln 2)$
$=\displaystyle \frac{1}{2}+\ln 2+\ln 3$
$=\displaystyle \frac{1}{2}+\ln 6$
