11TH•Mathematics•Partial Fractions•Question 2

Solution

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

Let:

\frac{2x+1}{(x-2)(x^2+3x+5)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+3x+5}

Multiply through:

2x+1=A(x^2+3x+5)+(Bx+C)(x-2)

Solving gives:

A=\frac13,\quad B=-\frac13,\quad C=\frac13

Hence:

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

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