Hey guys,

 

I could really use some hints on how to solve the integral [math] \int \frac{x}{1+\sqrt{x}} dx [/math]

I tried some substitutions, and doing some manipulations on the expression, with no solution. Any help would be much appreciated.

 

Regards,

 

Kerbox

Have you tried letting y=sqrt(x).

 

Then x=y^2

dx = 2ydy

 

then the integrand is all polynomials and should be much easier to integrate.

Let [math]x=u^2[/math], and do as Bignose said. The integral becomes [math]2\int \frac{u^3}{1+u} du[/math], which with some easy polynomial division will get you [math]\frac{u^3}{1+u}=u^2-u+1 -\frac{1}{1+u}[/math].

Split the integral in two, you get: [math]2(\int u^2 -u +1 du -\int \frac{1}{1+u} du[/math]. The first one is easy, reverse power rule. The second, do some substitution. So you should get [math]2(\frac{u^3}{3} - \frac{u^2}{2} + u -\log_e |u+1|)[/math].

 

Substitute [math]u=\sqrt {x}[/math] back in

 

EDIT: Btw I checked the answers, its correct