Question

Evaluate question 4 only​

Answer 1

Substitute $y = \sqrt x$, so that $y^2 = x$ and $2y\,dy = dx$. Then the integral becomes

$\displaystyle \int \frac{dx}{\sqrt{1 + \sqrt x}} = 2 \int \frac y{\sqrt{1+y}} \, dy$

Now substitute $z=1+y$, so $dz=dy$. The integral transforms to

$\displaystyle 2 \int \frac y{\sqrt{1+y}} \, dy = 2 \int \frac{z-1}{\sqrt z} \, dz = 2 \int \left(\sqrt z - \frac1{\sqrt z}\right) \, dz$

The rest is trivial. By the power rule,

$\displaystyle \int \left(\sqrt z - \frac1{\sqrt z}\right) \, dz = \frac23 z^{3/2} - 2z^{1/2} + C = \frac23 \sqrt z (z - 3) + C$

Put everything back in terms of $y$, then $x$ :

$\displaystyle 2 \int \frac y{\sqrt{1+y}} \, dy = \frac43 \sqrt{1+y} (y - 2) + C$

$\displaystyle \int \frac{dx}{\sqrt{1+\sqrt x}} = \boxed{\frac43 \sqrt{1+\sqrt x} (\sqrt x - 2) + C}$