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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!

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2  
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
4  
You haven't finished expressing the integrand in terms of u. –  Qiaochu Yuan Sep 15 '10 at 23:25
2  
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. Sep 15 '10 at 23:41
1  
@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 5:29

4 Answers 4

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 $u$.)

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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.

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

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How do you go on? –  Jack Aug 28 '11 at 14:03
2  
Surprisingly? ok, but useful? Sorry... –  draks ... Jul 14 '12 at 20:21

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

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.

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1  
You can write square root as $\sqrt{x}$: $\sqrt{x}$. More complex example: $\sqrt{1+\sqrt t}$ $\sqrt{1+\sqrt t}$. –  Martin Sleziak Apr 14 at 11:31

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