Kerbox Posted January 22, 2007 Share Posted January 22, 2007 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 Link to comment Share on other sites More sharing options...
Bignose Posted January 22, 2007 Share Posted January 22, 2007 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. Link to comment Share on other sites More sharing options...
Ragib Posted January 23, 2007 Share Posted January 23, 2007 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 Link to comment Share on other sites More sharing options...
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