Calculus question

Question

Integral of(x+1/x)^(1/2)

Answer 1

(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.