Part (1)

${\displaystyle \frac{10x^2-2x+4}{x^3+x}=\frac{A}{x}+\frac{\mathrm{Bx}+C}{x^2+1}.}$

In order to work out the constants $A,B,C$, I put the RHS of the identity over a common denominator:

${\displaystyle \frac{10x^2-2x+4}{x^3+x}=\frac{A(x^2+1)+x(\mathrm{Bx}+C)}{x(x^2+1)}.}$

Since the denominators are equal, it follows that the numerators of the fraction must also be equal:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{10x^2-2x+4}& =A(x^2+1)+x(\mathrm{Bx}+C)\\ \multicolumn{1}{c}{}& ={\mathrm{Ax}}^2+{\mathrm{Bx}}^2+\mathrm{Cx}+A\\ \multicolumn{1}{c}{}& =(A+B)x^2+\mathrm{Cx}+A\end{array}.}$

By comparing coefficients, it follows that A+B=10, C=-2 and A=4 so B=6.

Part (2)

${\displaystyle \frac{A}{x}+\frac{\mathrm{Bx}+C}{x^2+1}=\frac{A}{x}+\frac{D}{(x-i)}+\frac{E}{(x+i)}.}$

By taking $A/x$ away and replacing $B$ and $C$ by their respective values, and putting $D$ and $E$ over a common denominator:

${\displaystyle \frac{6x-2}{(x^2+1)}=\frac{D(x+i)+E(x-i)}{(x-i)(x+i)}.}$

Again, since the denominators are equal, it follows that the numerators are equal so $6x-2=D(x+i)+E(x-i)$.

[Editor's note: We need to take care here as the identity is not defined for $x=\pm i$ or $x=o$ (when the denominators of the fractions are zero).]

By comparing coefficients we have $D+E=6$ and $(D-E)i=-2$. Then $6i-2=D(2i)$ so $D=(6i-2)/2i$. Multiply both the numerator and the denominator by $i$ to get a real denominator:

${\displaystyle D=\frac{6i^2-2i}{-2}=\frac{-6-2i}{-2}=3+i.}$

Then $E(-2i)=-6i-2$, so $E=3-i$. So A=4, B=6, C=-2, D=3+i and E=3-i.

${\displaystyle \frac{10x^2-2x+4}{x^3+x}=\frac{4}{x}+\frac{6x-2}{x^2+1}=\frac{4}{x}+\frac{3+i}{(x-i)}+\frac{3-i}{(x+i)}.}$

[Editor's note: To be absolutely rigorous, since the expression is undefined for certain values of $x$, we should consider limits and we get the same result:

${\displaystyle D=\underset{x\to i}{lim}\frac{6x-2}{x+i}=\frac{6i-2}{2i}=3+i.}$

and similarly

${\displaystyle E=\underset{x\to -i}{lim}\frac{6x-2}{x-i}=\frac{-6i-2}{-2i}=3-i.]}$