Resolve $\dfrac{x^2+1}{(x-1)(x-2)(x-3)}$ into partial fractions.

Question

Accepted Answer

**Step 1.** Write
$$\frac{x^2+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}.$$

**Step 2.** Clear denominators:
$$x^2+1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2).$$

**Step 3.** Substitute $x=1$: $1+1=A(-1)(-2)\ \Rightarrow\ 2=2A\ \Rightarrow\ A=1$.

**Step 4.** Substitute $x=2$: $4+1=B(1)(-1)\ \Rightarrow\ 5=-B\ \Rightarrow\ B=-5$.

**Step 5.** Substitute $x=3$: $9+1=C(2)(1)\ \Rightarrow\ 10=2C\ \Rightarrow\ C=5$.

**Step 6 (check at $x=0$).** LHS $=\dfrac{1}{(-1)(-2)(-3)}=-\dfrac16$. RHS $=\dfrac1{-1}-\dfrac5{-2}+\dfrac5{-3}=-1+2.5-1.6\overline{6}=-\dfrac16$. They agree.

> [!ANSWER] $\dfrac{x^2+1}{(x-1)(x-2)(x-3)}=\dfrac1{x-1}-\dfrac5{x-2}+\dfrac5{x-3}$.

Q.Resolve $\dfrac{x^2+1}{(x-1)(x-2)(x-3)}$ into partial fractions.

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Q.Resolve $\dfrac{x^2+1}{(x-1)(x-2)(x-3)}$ into partial fractions.

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