Solution Let: 1x(x+1)3=Ax+Bx+1+C(x+1)2+D(x+1)3\frac{1}{x(x+1)^3}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}+\frac{D}{(x+1)^3}x(x+1)31=xA+x+1B+(x+1)2C+(x+1)3D Multiply by x(x+1)3x(x+1)^3x(x+1)3: 1=A(x+1)3+Bx(x+1)2+Cx(x+1)+Dx1=A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx1=A(x+1)3+Bx(x+1)2+Cx(x+1)+Dx Put x=0x=0x=0: 1=A⇒A=11=A\Rightarrow A=11=A⇒A=1 Put x=−1x=-1x=−1: 1=−D⇒D=−11=-D\Rightarrow D=-11=−D⇒D=−1 Solving remaining coefficients gives B=−1B=-1B=−1, C=−1C=-1C=−1. Hence: 1x(x+1)3=1x−1x+1−1(x+1)2−1(x+1)3\boxed{\frac{1}{x(x+1)^3}=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{(x+1)^2}-\frac{1}{(x+1)^3}}x(x+1)31=x1−x+11−(x+1)21−(x+1)31