∫dx(x+1)(x+5)Let,1(x+1)(x+5)=Ax+1+Bx+5....(1)1(x+1)(x+5)=(x+5)A+(x+1)B(x+1)(x+5)1=(x+5)A+(x+1)B1=Ax+5A+Bx+B1=(A+B)x+(5A+B)EquatingcoefficentsofxbothsideswegetA+B=0....(2)and5A+B=1.....(3)solvingaboveequation(2)and(3)wegetA=14andB=−14PutvaluesofAandBin(1)weget1(x+1)(x+5)=14(x+1)−14(x+5)Now,∫dx(x+1)(x+5)=∫dx4(x+1)−∫dx4(x+5)=14[ln(x+1)−ln(x+5)]=14ln(x+1x+5)