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

The quadratic factor $x^2+1$ has no real roots (it cannot be factorised further), so it needs a linear numerator $Bx+C$ rather than a single constant.

Write $\dfrac{2x+1}{(x-1)(x^2+1)} = \dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}$, giving $2x+1 = A(x^2+1)+(Bx+C)(x-1)$.

Since $x^2+1$ has no real linear factors, write $\dfrac{2x+1}{(x-1)(x^2+1)} = \dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}$. Multiplying both sides by $(x-1)(x^2+1)$:

Put $x=1$: $3 = A(2) \Rightarrow A=\dfrac{3}{2}$.

Expand the right side: $A x^2+A + Bx^2-Bx+Cx-C = (A+B)x^2 + (C-B)x + (A-C)$.

Comparing the coefficient of $x^2$ (which is $0$ on the left): $0 = A+B \Rightarrow B = -A = -\dfrac{3}{2}$.

Comparing the constant term: $1 = A - C \Rightarrow C = A-1 = \dfrac{3}{2}-1 = \dfrac{1}{2}$.

So $\dfrac{2x+1}{(x-1)(x^2+1)} = \dfrac{3/2}{x-1} + \dfrac{-\frac{3}{2}x+\frac{1}{2}}{x^2+1} = \dfrac{3}{2(x-1)} + \dfrac{1-3x}{2(x^2+1)}$.

Check (independent verification): compare the coefficient of $x$ on both sides as a cross-check not yet used above — left side has coefficient $2$; right side gives $C-B = \dfrac{1}{2}-\left(-\dfrac{3}{2}\right)=2$. This matches exactly, confirming $A$, $B$, $C$ are all correct without needing a further numeric substitution.

Writing the quadratic factor's numerator as a single constant $C$ instead of the linear form $Bx+C$ — a constant numerator cannot match every possible proper fraction over an irreducible quadratic, and the decomposition will fail to balance.