Answer
$\frac{1}{2}+\ln6$
Work Step by Step
$\int_1^2\frac{x^{3}+4x^{2}+x-1}{x^{3}+x^{2}}dx$ = $\int_1^2\left(1+\frac{3x^{2}+x-1}{x^{3}+x^{2}}\right)dx$ = $\int_1^2\left[1+\frac{3x^{2}+x-1}{x^2(x+1)}\right]dx$ = $\int_1^2\left[1+\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\right]dx$
$3x^{2}+x-1$ = $Ax(x+1)+B(x+1)+Cx^2$
$$\begin{cases}
A+C&=3\\
A+B&=1\\
B&=-1
\end{cases}$$
$A$ = $2$, $B$ = $-1$, $C$ = $1$
So
$\int_1^2\frac{x^{3}+4x^{2}+x+1}{x^{3}+x^{2}}dx$ = $\int_1^2\left(1+\frac{2}{x}-\frac{1}{x^2}+\frac{1}{x+1}\right)dx$ = $\left(x+2\ln|x|+\frac{1}{x}+\ln|x+1|\right)_1^2$ = $\left(2+2\ln2+\frac{1}{2}+\ln3\right)-(1+2\ln1+1+\ln2)$ = $\frac{1}{2}+\ln6$
