∫(x+1)dx/x(x+4)^(2)

Question

please answer this right now co'z i really need this thank you..

distinct quadratic factors 

of ∫(x+1)dx/x(x+4)^(2)

Answer 1

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.