Answer
$ \displaystyle \frac{3}{2}\ln(x^{2}+1)-3\tan^{-1}x+\sqrt{2}\tan^{-1}(\frac{x}{\sqrt{2}})+C$
Work Step by Step
The degree of the numerator is 3, the denominator's is 4. Try partial fractions.
The denominator has two prime quadratic factors.
$\displaystyle \frac{3x^{3}-x^{2}+6x-4}{(x^{2}+1)(x^{2}+2)}=\frac{Ax+B}{(x^{2}+1) }+\frac{Cx+D}{(x^{2}+2)}$
$3x^{3}-x^{2}+6x-4=(Ax+B)(x^{2}+2)+(Cx+D)(x^{2}+1) $
$3x^{3}-x^{2}+6x-4=(A+C)x^{3}+(B+D)x^{2}+(2A+C)x+(2B+D)$
Equating coefficients of $x^{3}$ and $x: \left\{\begin{array}{l}
A+C=3\\
2A+C=6
\end{array}\right.$
Subtract the first equation from the second: $A=3, C=0$
Equating coefficients of $x^{2}$ and the linear term$:$
$\left\{\begin{array}{l}
B+D=-1\\
2B+D=-4
\end{array}\right.\Rightarrow B=-3,D=2$
$I=\displaystyle \int\frac{3x^{3}-x^{2}+6x-4}{(x^{2}+1)(x^{2}+2)}dx= \int\frac{3x-3}{x^{2}+1 }dx+\int\frac{2}{x^{2}+2}dx$
$= \displaystyle \frac{3}{2}\int\frac{2x }{x^{2}+1 }dx-3\int\frac{dx}{x^{2}+1 }+2\int\frac{dx}{x^{2}+2}$
... first integral : $\displaystyle \left[\begin{array}{l}
t=x^{2}+1\\
dt=2xdx
\end{array}\right]\Rightarrow I_{1}\int\frac{dt}{t}=\ln|t|+C$
... for the second and third, table integral 17:
$\displaystyle \int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}+C$
$I==\displaystyle \frac{3}{2}\ln(x^{2}+1)-3\tan^{-1}x+2\cdot\frac{1}{\sqrt{2}}\tan^{-1}(\frac{x}{\sqrt{2}})+C$
$=\displaystyle \frac{3}{2}\ln(x^{2}+1)-3\tan^{-1}x+\sqrt{2}\tan^{-1}(\frac{x}{\sqrt{2}})+C$
