Integral of $\sqrt{1 + \sqrt{x}}$

Question

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!

Answer 1

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 $u$.)

Answer 2

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.

Answer 3

Suprisingly useful hint: $x = 1+(x-1)$.

Answer 4

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.

Answer 5

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) $n$.