Integral of $\sqrt{1 + \sqrt{x}}$ My professor wants us to do this problem to refresh ourselves with substitution. We have to solve:

$$\int\sqrt{1 + \sqrt{x}}\,\mathrm dx$$
  $$\int\sqrt{1 + \sqrt{1 + \sqrt{x}}}\,\mathrm dx$$
  ...etc...

If someone could point me in the right direction with the first one, I think I can handle the others.
Thanks for the help guys!
 A: $\int\sqrt{1 + \sqrt{x}}.dx$
t=1+$\sqrt{x}$
t-1 = $\sqrt{x}$
${(t-1)}^2$ = x
$\int\sqrt{t}.2(t-1)dt$
2$\int\sqrt{t}.(t-1)dt$
=2$\int\{t^\frac{3}{2}-t^\frac{1}{2})dt$
4$(\frac{t^\frac{5}{2}}{5}-\frac{t^\frac{3}{2}}{3})dt$
$\int\sqrt{1 + \sqrt{1+\sqrt{x}}}.dx$ -- It will be also resolved via Substitution Method used above.
A: One technique usually worth trying in integration is to be extremely optimistic: ask yourself what is "hard" about the integrand and then make a substitution that simplifies it, even if the substitution looks awful.  In your case, the "hard" part about $\sqrt{1 + \sqrt{x}}$ is that the argument of the outer root is not evidently a perfect square, so it doesn't simplify.  So substituting $u^2$ for $1 + \sqrt{x}$ would be worth considering.  (You will obtain an integrand that is a polynomial in $u$.)
A: This is the answer from Wolframalpha: note that you have to use the substitution
$$u=\sqrt{x}$$
first. Note that Wolframalpha able to give you the full step by step solution.
A: Suprisingly useful hint: $x = 1+(x-1)$.
A: Hint:
With $u=\sqrt{1+\sqrt x}$, you have $x=(u^2-1)^2.$
With $u=\sqrt{1+\sqrt{1+\sqrt x}}$, you have $x=((u^2-1)^2-1)^2.$
In both cases, a polynomial and the integral is easy. This generalizes.
A: Let $f(y):=\sqrt{1+y}$, with inverse $g(y):=y^2-1$. You want to evaluate$$\int f^{\circ n}(\sqrt{x})dx$$(the exponent with $\circ$ denoting repeated application of a function), where your first integral has $n=0$. With$$u=f^{\circ n}(\sqrt{x})\iff x=(g^{\circ n}(u))^2$$(where the ${}^2$ denotes squaring), integration by parts proves the result is$$xu-\int2xg^{\circ n}(u)\prod_{k=0}^{n-1}g^\prime(g^{\circ k}(u))du.$$The last integral is just a polynomial in $u$, albeit one we'd have to work out for a specific (hopefully small, or it'll take ages) $n$.
