Accedmychevron_right11thchevron_rightmathchevron_rightPartial Fractionschevron_rightExercise 5.1

Solution

Let:

2x+3(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3\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)2x+3=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)

Solving gives A=12A=\tfrac12, B=1B=1, C=32C=-\tfrac32.

2x+3(x+1)(x+2)(x+3)=12(x+1)+1x+232(x+3)\boxed{\frac{2x+3}{(x+1)(x+2)(x+3)}=\frac{1}{2(x+1)}+\frac{1}{x+2}-\frac{3}{2(x+3)}}