(1) ∫√{(x+1)/x}dx or (2) ∫√{(x+(1/x))}dx?
Please clarify!
(1) Let u2=x, then 2udu=dx.
∫[√(u2+1)/u]2udu=2∫√(u2+1)du;
let u=tanθ, u2+1=sec2θ; du=sec2θdθ.
2∫√(u2+1)du=2∫sec3θdθ.
Let J=∫sec3θdθ.
J=∫secθ(1+tan2θ)dθ=∫secθdθ+∫secθtan2θdθ.
∫secθdθ=∫secθ(secθ+tanθ)dθ/(secθ+tanθ)=∫(secθtanθ+sec2θ)dθ/(secθ+tanθ)=
ln|secθ+tanθ|.
J=ln|secθ+tanθ|+∫secθtan2θdθ.
Let K=∫secθtan2θdθ; let p=tanθ, dp=sec2θdθ; let dq=secθtanθdθ, q=secθ.
∫pdq=∫tanθsecθtanθdθ=K=pq-∫qdp=secθtanθ-∫sec3θdθ=secθtanθ-J.
So J=ln|secθ+tanθ|+K=ln|secθ+tanθ|+secθtanθ-J, 2J=ln|secθ+tanθ|+secθtanθ.
2∫√(u2+1)du=2∫sec3θdθ=2J=ln|secθ+tanθ|+secθtanθ; sec2θ=1+tan2θ=u2+1; secθ=√(u2+1).
2∫√(u2+1)du=ln|√(u2+1)+u|+u√(u2+1)=ln(√(x+1)+√x)+√(x(x+1)).
∫√{(x+1)/x}dx=ln(√(x+1)+√x)+√(x(x+1))+C, where C is integration constant. (This seems to be the more likely interpretation of the question.)
CHECK
Differentiate ln(√(x+1)+√x)+√(x(x+1))+C:
{(1/(√(x+1)+√x))(1/[2√(x+1)]+1/(2√x)}+{(2x+1)/√(2x(x+1))}=
{½((√x+√(x+1))/(√(x+1)+√x))(1/√(x(x+1)))}+{½(2x+1)/√(x(x+1))}=
½(1/√(x(x+1)))+(2x+1)/√(x(x+1)))=½((2x+2)/√(x(x+1)))=√(x+1)/√x=√((x+1)/x), which is the given integrand✔️
(2) Solution to this interpretation may involve expressing the integrand as an infinite series, then applying integration to each term in the series to obtain a general solution.