I=∫{(x+1)/[x(x+4)2]}dx.
Let (x+1)/[x(x+4)2]=A/x+B/(x+4)+Cx/(x+4)2
x+1=A(x+4)2+Bx(x+4)+Cx2
x+1=Ax2+8Ax+16A+Bx2+4Bx+Cx2
Matching coefficients:
constant: 16A=1, A=1/16
x: 8A+4B=1, B=⅛
x2: A+B+C=0, C=-3/16.
I=(1/16)[∫dx/x+2∫dx/(x+4)-3∫(x/(x+4)2)dx].
x/(x+4)2=(x+4-4)/(x+4)2=1/(x+4)-4/(x+4)2.
I=(1/16)[∫dx/x+2∫dx/(x+4)-3∫(dx/(x+4)+12∫dx/(x+4)2],
I=(1/16)[∫dx/x-∫dx/(x+4)+12∫dx/(x+4)2],
I=(1/16)[ln|x|-ln|x+4|-12/(x+4)]=ln|x/(x+4)|/16-3/(4(x+4))+C, where C is integration constant.