Q.$\dfrac{x}{(x^2+1)(x-1)(x+2)}$

Step 1. Write $\dfrac{x}{(x^2+1)(x-1)(x+2)}=\dfrac A{x-1}+\dfrac B{x+2}+\dfrac{Cx+D}{x^2+1}$, so $x=A(x+2)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+2)$.

Step 2. $x=1$: $1=A(3)(2)=6A\Rightarrow A=\dfrac16$. $x=-2$: $-2=B(-3)(5)=-15B\Rightarrow B=\dfrac2{15}$.

Step 3. $x=0$: $0=A(2)(1)+B(-1)(1)+D(-1)(2)=2A-B-2D$. With $2A=\dfrac13,\ B=\dfrac2{15}$: $\dfrac13-\dfrac2{15}-2D=0\Rightarrow\dfrac3{15}=2D\Rightarrow D=\dfrac1{10}$.

Step 4. $x=2$: $2=A(4)(5)+B(1)(5)+(2C+D)(1)(4)=20A+5B+4(2C+D)$. $20A=\dfrac{10}3,\ 5B=\dfrac23$, sum $=4$. So $2=4+4(2C+D)\Rightarrow2C+D=-\dfrac12\Rightarrow2C=-\dfrac12-\dfrac1{10}=-\dfrac35\Rightarrow C=-\dfrac3{10}$.

Cover-up rule for the two linear factors, then substitute two convenient extra values (e.g. $x=0,2$) to pin the quadratic-factor constants

• Sign error combining fractions when solving for $D$.
• Forgetting the numerator over the irreducible quadratic factor must be linear ($Cx+D$), not just a constant.