## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{1}{x}-\dfrac{x}{x^2+1}$
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: $$\displaystyle \frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{(-1)x+0}{x^2+1}\\ \displaystyle \frac{1}{x(x^2+1)}=\frac{1}{x}-\frac{x}{x^2+1}$$