int x/sqrt(6x+1) dx Find the indefinite integral

intx/sqrt(6x+1)dx
Apply integral substitution: u=6x+1
=>(u-1)=6x
=>x=(u-1)/6
dx=1/6(du)
intx/sqrt(6x+1)dx=int((u-1)/6)/sqrt(u)(1/6)du
=int1/36(u-1)/(sqrt(u))du
Take the constant out,
=1/36int(u-1)/sqrt(u)du
=1/36int(u/sqrt(u)-1/sqrt(u))du
=1/36int(u^(1/2)-u^(-1/2))du
Apply the sum rule,
=1/36{intu^(1/2)du-intu^(-1/2)du}
Apply the power rule,
=1/36{(u^(1/2+1)/(1/2+1))-(u^(-1/2+1)/(-1/2+1))}
=1/36{u^(3/2)/(3/2)-u^(1/2)/(1/2)}
=1/36{2/3u^(3/2)-2u^(1/2)}
Substitute back u=(6x+1) and add a constant C to the solution,
=1/36(2/3(6x+1)^(3/2)-2(6x+1)^(1/2))+C
=1/36(2)(6x+1)^(1/2)(1/3(6x+1)-1)+C
=1/18sqrt(6x+1)((6x+1-3)/3)+C
=sqrt(6x+1)/18((6x-2)/3)+C
=sqrt(6x+1)/18(2/3)(3x-1)+C
=1/27(3x-1)sqrt(6x+1)+C