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

Solution 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 $u$.)

Solution 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.

Solution 3:

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.