I think what you came up is correct assuming all the given is correct. To proceed from there, say that n=1+u, so that dn=du. Since n=1+u, we can say that u=n-1. You should come up with integral of (n-1)/n dn, so that if you simplify, you get integral of 1-1/n dn. If you integrate that you get 2[n-ln(abs(n))] + c. Sub back then you get 2(1+u)-2ln(abs(1+u))+c. Sub back again, you get 2(1+sqrt(x-1))-2ln(abs(1+sqrt(x-1)))+c. Im not sure though if you would need that Plus or minus from the root. By the way, sorry for not using latex, I'm kinda on a rush.