Answer
$\displaystyle \frac{1}{3}\ln|x^{3}+1|-\frac{1}{3}\ln|x^{3}+2|+C$
Work Step by Step
$I=\displaystyle \int\frac{x^{2}}{x^{6}+3x^{3}+2}dx=\quad \left[\begin{array}{ll}
u=x^{3} & \\
du=3x^{2}dx & x^{2}dx=\frac{du}{3}
\end{array}\right]$
$=\displaystyle \frac{1}{3}\int\frac{du}{u^{2}+3u+2}$
factor the denominator. (2)(1)=2, $\quad$ (2+1)=3....
$=\displaystyle \frac{1}{3}\int\frac{du}{(u+1)(u+2)}$
... use the partial fraction method
$\displaystyle \frac{1}{(u+1)(u+2)}$=$\displaystyle \frac{A}{u+1}$+$\displaystyle \frac{B}{u+2}$
$1=Au+2A+Bu+B$
$1=(A+B)u+(2A+B)\Rightarrow\left\{\begin{array}{l}
A+B=0\\
2A+B=1
\end{array}\right.$
Subtract the first from the second equation: $A=1, B=-1.$
$I=\displaystyle \frac{1}{3}\int\frac{du}{u+1}-\frac{1}{3}\int\frac{du}{u+2}$
Both are type 2 integrals (from the table)
$=\displaystyle \frac{1}{3}\ln|u+1|-\frac{1}{3}\ln|u+2|+C$
... bring back x
$=\displaystyle \frac{1}{3}\ln|x^{3}+1|-\frac{1}{3}\ln|x^{3}+2|+C$
