Answer
$\frac{1}{x-2}-\frac{x+1}{x^2+4}$
Work Step by Step
Step 1. Factorize the denominator as: $x^2(x-2)+4(x-2)=(x-2)(x^2+4))$
Step 2. Assume the partial fraction decomposition is:
$\frac{A}{x-2}+\frac{Bx+C}{x^2+4}$
Step 3. Combine the rational functions above:
$\frac{A(x^2+4)+(Bx+C)(x-2)}{(x-2)(x^2+4)}\frac{(A+B)x^2+(-2B+C)x+(4A-2C)}{x(x-1)^2}$
Step 4. Compare the above function with the original to setup the following system of equations:
$\begin{cases} A+B=0\\-2B+C=1\\4A-2C=6 \end{cases}$
Step 5. Solve the above equations: equation3 gives $2A-C=3$ add it to equation2 gives $2A-2B=4$ or $A-B=2$, combine with equation1 to get $A=1, B=-1$, thus $C=-1$
Step 6. The final result is:
$\frac{1}{x-2}-\frac{x+1}{x^2+4}$
