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

Split into $\dfrac{A}{x+2}+\dfrac{Bx+C}{x^2+1}$, clear denominators, and match coefficients (or substitute the root of the linear factor) to find $A,B,C$.

Assume

$\frac{x^2-3}{(x+2)(x^2+1)} = \frac{A}{x+2}+\frac{Bx+C}{x^2+1}.$

Multiplying both sides by $(x+2)(x^2+1)$:

$x^2-3 = A(x^2+1)+(Bx+C)(x+2).$

Find $A$: substitute $x=-2$ (which kills the second term):

$(-2)^2-3 = A\big((-2)^2+1\big) \Rightarrow 1 = 5A \Rightarrow A=\frac15.$

Find $B,C$: expand the right side: $Ax^2+A+Bx^2+2Bx+Cx+2C = (A+B)x^2+(2B+C)x+(A+2C)$.

Matching with $x^2+0\cdot x-3$:

$A+B=1 \Rightarrow B=1-\frac15=\frac45,$

$2B+C=0 \Rightarrow C=-2B=-\frac85.$

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