Question

integrate √1-√x/√1+√x

Answer 1

I = ( 1 - √x)dx/( 1 + √x )
I = ( 1 -√x)(1 -√x)dx/(1 - x)
I = ( 1 + x - 2√x)dx/(1 - x)
I = ( 1 - x + 2x - 2√x)dx/(1 - x)
I = ( 1 - x)dx/(1 - x) - 2√x( 1 - √x)/(1 + √x)(1 -√x)
I = 1.dx - 2√x.dx/(1 + √x)
I = x - P

P = 2√x.dx/(1 + √x)
Let √x = Z
dx/2√x = dZ

P = 4Z²dZ/(1 + Z)
=4{ Z + 1/( Z + 1) - 1 }dZ
= 4Z²/2 + 4ln(1 + Z) -4Z
= 2Z² + 4ln(1 + Z) -4Z
= 2x + 4ln(1 + √x) -4√x

now,

I = x - 2x - 4ln(1 + √x) + 4√x
I = - x - 4ln(1 + √x) + 4√x