11TH•Mathematics•Partial Fractions•Question 4

Solution

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

Let:

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

Then:

2x+3=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)

Solving gives $A=\tfrac12$ , $B=1$ , $C=-\tfrac32$ .

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

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