Answer
$\dfrac{1}{x}-\dfrac{x}{x^2+1}$
Work Step by Step
We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has only non-repeated irreducible quadratic factor.
So, we will write the given rational expression into partial fraction form as:
$$\displaystyle \frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1} \quad \quad(1)$$
Now, write the RHS with a common denominator.
\begin{align*}
\frac{1}{x(x^2+1)}&=\frac{A(x^2+1)+(Bx+C)(x)}{x(x^2+1)}\\
\\\frac{1}{x(x^2+1)}&=\frac{Ax^2+A+Bx^2+Cx}{x(x^2+1)}\end{align*}
Next, equate the numerators to obtain:
$$Ax^2+A+Bx^2+Cx=1 \\
\\(Ax^2+Bx^2)+Cx+A=1 \\
(A+B)x^2+Cx+A=1$$
Equate the coefficients of the polynomials on the LHS and RHS to obtain:
$$A+B=\quad \quad(2)
\\C=0 \quad \quad(3)
\\A=1\quad \quad4)$$
Plug $A=1$ into the Equation $(2)$ to solve for $B$.
$$1+B=0 \implies B=-1$$
Thus, the equation $(1)$ becomes:
$$\displaystyle \frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{(-1)x+0}{x^2+1}\\
\displaystyle \frac{1}{x(x^2+1)}=\frac{1}{x}-\frac{x}{x^2+1}$$
