int 3/(2sqrt(x)(1+x)) dx Find the indefinite integral

For the given integral: int 3/(2sqrt(x)(1+x)) dx , we may apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int 3/(2sqrt(x)(1+x)) dx = 3/2int 1/(sqrt(x)(1+x)) dx .
 
For the integral part, we apply u-substitution by letting:
u = sqrt(x)
We square both sides to get: u^2 = x .
Then apply implicit differentiation, we take the derivative on both sides with respect to x as:
2u du =dx .
Plug-in dx= 2u du , u =sqrt(x) and x= u^2 in the integral:
3/2int 1/(sqrt(x)(1+x)) dx =3/2int 1/(u(1+u^2)) (2u du)
Simplify by cancelling out u and 2 from top and bottom:
3/2int 1/(u(1+u^2)) (2u du) =3 int 1/(1+u^2) du
The integral part resembles the basic integration formula  for inverse tangent:
int 1/(1+u^2) du = arctan (u) +C
 then, 
3 int 1/(1+u^2)  du = 3 * arctan(u) +C
Express in terms x by plug-in  u =sqrt(x) :
3 arctan(u) +C =3 arctan(sqrt(x)) +C
Final answer:
int 3/(2sqrt(x)(1+x)) dx = 3arctan(sqrt(x))+C