Evaluate question 4 only

Question

asked Apr 25, 2023 224k views

1 Answer

Sikandar Amla · Answer 1 · 2023-04-30T05:51:27+0000

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

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

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

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

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
, then
:

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

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