Question

S=Integral xdx/sqrt(x-1). I have proceeded thus-
Put sqrt(x-1)=u then x=u^2+1 and
dx/sqrt(x-1)=2du.
S=(u^2+1)2du/u
=(2u+2/u)du=u^2+2 log u +C
=(x-1)+ 2 log sqrt(x-1)=(x-1)+log(x-1)+C

Required answer is 2/3*(x+2)sqrt(x-1)+C
Have I proceeded correctly and if so can these 2 results be shown to be same except for constants of integration.
If not, where have I gone wrong?

Answer 1

Integral of x / sqrt(x-1) dx
We can use substitution method.
Let u = x - 1
Thus, x = u + 1 and dx = du
Substituting,
Integral of (u+1)/sqrt(u) du
Integral of (u+1)(u^(-1/2)) du
Integral of u^(1/2) + u^(-1/2) du
= (2/3)*u^(3/2) + (2)*u^(1/2) + C
= (2/3)*(x-1)^(3/2) + 2(x-1)^(1/2) + C
and if you simplify this, you'll get the answer you typed.

hope this helps :3