integration of this?

Question

1 / 2 (x)^1/2 [ 1 + (x)^1/2 ] dx

Answer 1

∫ (1 / [2 x^(1/2) [1 + x^(1/2)] dx)

First, I'm going to rearrange this in a form that makes the next step obvious.

∫ (1/[1 + x^(1/2)] 1/[2x^(1/2)] dx

Here's where we use substitution.

Let u = 1 + x^(1/2). Then
du = (1/2)x^(-1/2) dx, so
du = 1/[2x^(1/2)] dx

Note that 1/[2x^(1/2)] dx is the tail end of our integral, which means du will be the tail end after the substitution.

∫ (1/u) du

And this is now trivial to integrate.

ln|u| + C

But u = 1 + x^(1/2), so our final answer is

ln|1 + x^(1/2)| + C

Answer 2

No brackets used in question so assume that question is:-
I = (1/2) ∫ x^(1/2) dx + (1/2) ∫ x dx
I = (1/2) x^(3/2).(2/3) +(1/2) x² / 2 + C
I = (1/3).x^(3/2) + (1/4).x² + C

Answer 3

1 / 2 (x)^1/2 [ 1 + (x)^1/2 ] dx = 0.5(x^0.5 + x)dx

On integrating

0.5 (2/3x^3/2) + x^2/2) + C

=1/3x^(3/2) + 1/4x^2 + C