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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!

Martin Sleziak
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Swift
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    For the first one you want to try u = 1 + sqrt(x). – Qiaochu Yuan Sep 15 '10 at 22:39
  • Then I get 2 * Integral of sqrt(u)*sqrt(x), how do I find the anti derivative of that? – Swift Sep 15 '10 at 23:04
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    You haven't finished expressing the integrand in terms of u. – Qiaochu Yuan Sep 15 '10 at 23:25
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    i.e. after doing the substitution Qiaochu suggested, make sure you've gotten rid of all the $x$'s. Alternatively, you can also try the substitution $x=\tan^4;\theta$ and make sure everything is in terms of $\theta$ – J. M. ain't a mathematician Sep 15 '10 at 23:41
  • @J.M.: Your suggestion is great, but how did you think of it? – whuber Sep 16 '10 at 02:13
  • @whuber: $1+\tan^2 \theta=\mathrm{sec}^2 \theta$ . I dunno, it was the first thing to spring to mind upon seeing the question. – J. M. ain't a mathematician Sep 16 '10 at 02:26
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    @J.M.: I figure Mike is looking for principles or techniques more than answers, so I just wondered how you would articulate your insight. Generally, problems with roots often suggest trigonometric substitutions because of the quadratic identities enjoyed by circular functions. But suggesting the fourth power of the tangent was really taking two steps at once. (That's not a criticism, just an observation.) – whuber Sep 16 '10 at 05:29
  • As far as I know, substitutions do not help (and are not needed) to solve the other integrals beyond the first one. – T.. Sep 16 '10 at 05:30

5 Answers5

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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 $u$.)

whuber
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3

$\int\sqrt{1 + \sqrt{x}}.dx$

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.

user3487861
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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.

0

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) $n$.

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

T..
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