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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!

  • 2
    For the first one you want to try u = 1 + sqrt(x).2010-09-15
  • 0
    Then I get 2 * Integral of sqrt(u)*sqrt(x), how do I find the anti derivative of that?2010-09-15
  • 4
    You haven't finished expressing the integrand in terms of u.2010-09-15
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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$2010-09-15
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    @J.M.: Your suggestion is great, but how did you think of it?2010-09-16
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    @whuber: $1+\tan^2 \theta=\mathrm{sec}^2 \theta$ . I dunno, it was the first thing to spring to mind upon seeing the question.2010-09-16
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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.)2010-09-16
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    As far as I know, substitutions do not help (and are not needed) to solve the other integrals beyond the first one.2010-09-16

4 Answers 4

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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 $u$.)

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

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.

  • 1
    You can write square root as `$\sqrt{x}$`: $\sqrt{x}$. More complex example: `$\sqrt{1+\sqrt t}$` $\sqrt{1+\sqrt t}$.2014-04-14
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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: $x = 1+(x-1)$.

  • 0
    How do you go on?2011-08-28
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    *Surprisingly*? ok, but *useful*? Sorry...2012-07-14