Answer
$$\frac{1}{3}\ln |x^2+1|- \frac{1}{3}\ln |x^2+2|+C$$
Work Step by Step
Given
$$\int \frac{x^2}{x^6+3 x^3+2} d x$$
Since
\begin{aligned}
\frac{x^2}{x^6+3 x^3+2} &= \frac{x^2}{(x^3+1)(x^3+2)}
\end{aligned}
Let
$$ u=x^3+1\ \to \ \ du =3x^2dx$$
Then
\begin{aligned}
\int \frac{x^2dx}{x^6+3 x^3+2} &=\frac{1}{3}\int \frac{du}{(u)(u+1)}
\end{aligned}
Using partial fractions
\begin{aligned}
\frac{1}{u\left(u+1\right)}&=\frac{a_0}{u}+\frac{a_1}{u+1}\\
&=\frac{ a_0\left(u+1\right)+a_1u}{u(u+1)}\\
1&=a_0\left(u+1\right)+a_1u
\end{aligned}
at $ u=0 \ \to \ a_0= 1, \ \ \ \ u=-1\ \to \ \ a_1=-1$
Hence
\begin{aligned}
\int \frac{x^2dx}{x^6+3 x^3+2} &=\frac{1}{3}\int \frac{du}{(u)(u+1)} \\
&=\frac{1}{3}\int \frac{1}{u}du-\frac{1}{3}\int \frac{1}{u+1} du\\
&= \frac{1}{3}\ln |u|- \frac{1}{3}\ln |u+1|+C\\
&= \frac{1}{3}\ln |x^2+1|- \frac{1}{3}\ln |x^2+2|+C
\end{aligned}
