Answer
$\frac{1}{2}+\ln{\frac{27}{32}}$
Work Step by Step
$\int_0^1\frac{x^{2}+x+1}{(x+1)^2(x+2)}dx$ = $\int_0^1\left[\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}\right]dx$
$x^{2}+x+1$ = $A(x+1)(x+2)+B(x+2)+C(x+1)^2$
$x^{2}+x+1$ = $A(x^{2}+3x+2)+B(x+2)+C(x^{2}+2x+1)$
$1$ = $A+C$
$1$ = $3A+B+2C$
$1$ = $2A+2B+C$
$B$ = $1$, $A$ = $-2$, $C$ = $3$
So
$\int_0^1\frac{x^{2}+x+1}{(x+1)^2(x+2)}dx$ = $\int_0^1\left[-\frac{2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}\right]dx$ = $[-2\ln|x+1|-\frac{1}{x+1}+3\ln|x+2|]_0^1$ = $(-2\ln2-\frac{1}{2}+3\ln3)-(-2\ln1-1+3\ln2)$ = $\frac{1}{2}-5\ln2+3\ln3$ = $\frac{1}{2}+\ln{\frac{27}{32}}$
