`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`