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`
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