11TH•Mathematics•Partial Fractions•Question 6

Solution

Resolve into partial fractions: $\dfrac{4x^3+5x^2-3x-2}{x^2-1}$

\frac{4x^3+5x^2-3x-2}{x^2-1} =\frac{4x^3+5x^2-3x-2}{(x-1)(x+1)}

Divide first:

\frac{4x^3+5x^2-3x-2}{x^2-1}=4x+5+\frac{2x+3}{x^2-1}

Now:

\frac{2x+3}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}

Solving gives $A=2$ , $B=-1$ .

Hence:

\boxed{\frac{4x^3+5x^2-3x-2}{x^2-1}=4x+5+\frac{2}{x-1}-\frac{1}{x+1}}

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