Answer
$\displaystyle \dfrac{x+1}{x^2(x-2)}=\dfrac{\frac{-3}{4}}{x}+\dfrac{\frac{-1}{2}}{x^2}+\dfrac{\frac{3}{4}}{x-2}$
Work Step by Step
We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has non-repeated irreducible quadratic factor. So, we will write the given rational expression into partial fraction form as: $$\displaystyle \frac{x+1}{x^2(x-2)}=\frac{A}{x}+\frac{B}{x^2}+\dfrac{C}{x-2} …(1)$$
Now, write the RHS with a common denominator.
$$\displaystyle \frac{x+1}{x^2(x-2)}=\dfrac{Ax(x-2)+B(x-2)+Cx^2}{x^2(x-2)}\\ \displaystyle \frac{x+1}{x^2(x-2)}=\dfrac{Ax^2-2Ax+Bx-2B+Cx^2}{x^2(x-2)} \\ \displaystyle \frac{x+1}{x^2(x-2)}=\dfrac{(A+C)x^2+(-2A+B)x+(-2B)}{x^2(x-2)}$$
Next, equate the numerators to obtain:
$$(A+C)x^2+(-2A+B)x+(-2B)=x+1$$
Equate the coefficients of the polynomials on the LHS and RHS to obtain:
$$A+C=0 ~~~(2) \\ -2A+B=1 ~~~(3) \\ -2B=1~~~(4)$$
Equation (4) yields: $$B=\dfrac{-1}{2}$$
Substitute $B=\dfrac{-1}{2}$ into equations (3) to obtain: $$-2A-\dfrac{-1}{2}=1\\ -2A=1+\dfrac{1}{2}\\-2A=\dfrac{3}{2}\\ A=\dfrac{-3}{4}$$
Finally, substitute $A=\dfrac{-3}{4}$ into equations (2) to solve for $C$ $$-\dfrac{3}{4}+C=0 \\C=\dfrac{3}{4}$$
Therefore, the equation (1) becomes:
$$\displaystyle \dfrac{x+1}{x^2(x-2)}=\dfrac{\frac{-3}{4}}{x}+\dfrac{\frac{-1}{2}}{x^2}+\dfrac{\frac{3}{4}}{x-2}$$
