Answer
$\frac{x+1}{x^2+3}-\frac{1}{x}$
Work Step by Step
Step 1. Factor the denominator as: $x(x^2+3)$
Step 2. Assume the end results as: $\frac{A}{x}+\frac{Bx+C}{x^2+3}$
Step 3. Combine the above functions as:
$\frac{A(x^2+3)+(Bx+C)x}{x(x^2+3)}=\frac{(A+B)x^2+(C)x+(3A)}{x(x^2+3)}$
Step 4. Compare the above result with the original expression to set up the following system of equations:
\begin{cases} A+B=0 \\ C=1\\3A=-3 \end{cases}
Step 5. Solve the above equations to get $A=-1, B=1,C=1$
Step 6. Write the final results as:
$-\frac{1}{x}+\frac{x+1}{x^2+3}$ or $\frac{x+1}{x^2+3}-\frac{1}{x}$
