Integral check $\int \frac{6x+4}{x^4+3x^2+5} \ \text{dx}$ $$\int \frac{6x+4}{x^4+3x^2+5} \ \text{dx}$$ Partial fraction decomposition of  $\frac{6x+4}{x^4+3x^2+5}$ is of the following form: $$\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+1}$$ We need to find  $A,B,C$ and $D$ \ $$\frac{6x+4}{x^4+3x^2+5}=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+1}$$ Or, $$6x+4=x^3(A+C)+x^2(B+D)+x(2A+C)+2B+D$$  By solving: $$\begin{cases} A+C=0 \\ B+D=0 \\ 2A+C=6 \\2B+D=4 \end{cases}$$ We get $$A=-6,B=-4,C=6,D=4$$ Hence $$\frac{6x+4}{x^4+3x^2+5}=-\frac{6x+4}{x^2+2}+\frac{6x+4}{x^2+1}$$ Now, by linearity, $$\int \frac{6x+4}{x^4+3x^2+5} \ \text{dx}=-\int\frac{6x+4}{x^2+2}\ \text{dx}+\int\frac{6x+4}{x^2+1}\ \text{dx}$$ Or, 
$$-3\int\frac{2x}{x^2+2}\ \text{dx}-4\int\frac{1}{x^2+2}\ \text{dx}+3\int\frac{2x}{x^2+1}\ \text{dx}+4\int\frac{1}{x^2+1}\ \text{dx}$$ Which equals to:
$$-3\ln(x^2+2)+\frac{4}{\sqrt{2}}\text{arctan}(\frac{x}{\sqrt{2}})+3\ln(x^2+1)+4\text{arctan}(x)+C$$

All the signs of all your letters are switched. This is because $A+A+C=A+0=A=6$, $B+B+D=B+0=B=4$, etc. Also, $$(x^2+1)(x^2+2)=x^4+3x^3+2\neq x^2+3x+5$$ Your polynomial has no real roots, so you'll have to do some manipulation in $\Bbb C$, as follows:
Let $x^2=u$. Then $$u^2+3u+5=0$$ has solutions $$u_1=\frac{-3+i\sqrt 11}{2}$$
$$u_2=\frac{-3-i\sqrt 11}{2}$$
Thus $$u^2+3u+5=\left(u-\frac{-3-i\sqrt 11}{2}\right)\left(u-\frac{-3+i\sqrt 11}{2}\right)$$
We can write then $${u^2} + 3u + 5 = \left( {u + {3 \over 2} + {{i\sqrt 1 1} \over 2}} \right)\left( {u + {3 \over 2} - {{i\sqrt 1 1} \over 2}} \right)$$ and using the difference of squares factorization $${u^2} + 3u + 5 = {\left( {u + {3 \over 2}} \right)^2} - {{{i^2}11} \over 4} = {\left( {u + {3 \over 2}} \right)^2} + {{11} \over 4}$$ Thus $${x^4} + 3{x^2} + 5 = {\left( {{x^2} + {3 \over 2}} \right)^2} + {{11} \over 4}$$
You can move on from that.
