Answer
\[\int _0^1\frac{x^2+x+1}{(x+1)^2(x+2)}dx = \ln(\frac{27}{32})+\frac{1}{2}\]
Work Step by Step
Make \[\frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2} \]
Thus\[ \frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{(A+C)x^3+(4A+B+3C)x^2+(5A+3B+3C)x+2A+2B+C}{(x+1)^2(x+2)}\]
$\therefore$ \[
\begin{cases}
A=3\\
B=-2\\
C=1\\
\end{cases} \]
$\therefore$\[\int \frac{x^2+x+1}{(x+1)^2(x+2)}dx=\int [\frac{3}{x+2}-\frac{2}{x+1}+\frac{1}{(x+1)^2}]dx=3\ln|x+2|-2\ln |x+1|-\frac{1}{x+1}+C\]
\[\int _0^1\frac{x^2+x+1}{(x+1)^2(x+2)}dx = \ln(\frac{27}{32})+\frac{1}{2}\]
