After using #u=sqrtx#, #x=u^2# and #dx=2u*du# transforms, this integral became
#int (2usqrt(u^2-u))/(u^2-1)*du#
After using #sqrt(u^2-u)=(u-1)*y# transform,
#u^2-u=(u-1)^2*y^2#
#u*(u*1)=(u-1)^2*y^2#
#u=(u-1)*y^2#
#u=uy^2-y^2#
#y^2=uy^2-u#
#u=y^2/(y^2-1)#
#du=[2y*(y^2-1)-2y*y^2]/(y^2-1)^2*dy=(-2y*dy)/(y^2-1)^2#
Also denominator became,
#u^2-1=[y^2/(y^2-1)]^2-1=(2y^2-1)/(y^2-1)^2#
Also numerator became,
#2usqrt(u^2-u)=(2y^2)/(y^2-1)*[y^2/(y^2-1)-1]*y#
=#(2y^3)/(y^2-1)^2#
Hence,
#int (2y^3)/(y^2-1)^2*(-2y*dy)/(y^2-1)^2*(y^2-1)^2/(2y^2-1)]#
=#int (-4y^4*dy)/[(y^2-1)^2*(2y^2-1)]#
=#int (-4y^4*dy)/[(y+1)^2*(y-1)^2*(2y^2-1)]#
=#int (-4dy)/(2y^2-1)#-#int dy/(y+1)#-#int dx/(y+1)^2#-#int dy/(y-1)^2#+#int dy/(y-1)#
=#1/(y+1)+1/(y-1)+ln(y-1)-Ln(y+1)-int (8dy)/(4y^2-2)#
=#(2y)/(y^2-1)+ln((y-1)/(y+1))-2sqrt2int (2sqrt2dy)/(4y^2-2)#
=#(2y)/(y^2-1)+ln((y-1)/(y+1))-2sqrt2int (dy)/(2y-sqrt2)+2sqrt2int (dy)/(2y+sqrt2)#
=#(2y)/(y^2-1)+ln((y-1)/(y+1))-sqrt2ln(2y-sqrt2)+sqrt2ln(2y+sqrt2)+C#
=#(2y)/(y^2-1)+ln((y-1)/(y+1))+sqrt2Ln((2y+sqrt2)/(2y-sqrt2))+C#
=#(2y)/(y^2-1)+ln((y-1)/(y+1))+sqrt2Ln((ysqrt2+1)/(ysqrt2-1))+C#
After using #y=sqrt(u^2-u)/(u-1)# inverse transforms, it became
#2sqrt(u^2-u)+ln((sqrt(u^2-u)-u+1)/(sqrt(u^2-u)+u-1))+sqrt2Ln((sqrt(2u^2-2u)+u-1)/(sqrt(2u^2-2u)-u+1))+C#
After using #u=sqrtx# inverse transforms, I found
#sqrt(x-sqrtx)/(x-1)*dx#
=#2sqrt(x-sqrtx)#+#ln((sqrt(x-sqrtx)-sqrtx+1)/(sqrt(x-sqrtx)+sqrtx-1))#+#sqrt2Ln((sqrt(2x-2sqrtx)+sqrtx-1)/(sqrt(2x-2sqrtx)-sqrtx+1))+C#
