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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!

  • 0
    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 $u$.)

2

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

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
1

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.

0

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

  • 4
    *Surprisingly*? ok, but *useful*? Sorry...2012-07-14