Integration by substitution, why is the $u$ this value? $$\int\frac{1}{(x+3)\sqrt{x}}~dx$$
I was wondering how you integrate this. I know you use substitution however I think of using $\sqrt x~$ for $u$, however on the integration calculator it says to use $u = \dfrac{\sqrt x}{\sqrt 3}~.$ I don't understand why and what to do once that is the substitution value. 
Please help 

Let $x=u^2$ so that $dx=2u\cdot du$. Then your integral is transformed into $$=\int\frac{2u\cdot du}{(u^2+3)\sqrt{u^2}}=2\int\frac{du}{u^2+3}$$
where the latter can be handled as an arctangent integral. 
To flesh it out completely, $$=2\int\frac{du}{u^2+3}=2\int\frac{du}{3\left(\frac{u^2}{3}+1\right)}=\frac{2}{3}\int\frac{du}{\left(\frac{u}{\sqrt{3}}\right)^2+1}$$ with $t=\frac{u}{\sqrt{3}}\implies dt=\frac{du}{\sqrt{3}}\implies\sqrt{3}dt=du\implies $ $$=\frac{2}{3}\int\frac{\sqrt{3}}{t^2+1}dt=\frac{2\sqrt{3}}{3}\int\frac{dt}{t^2+1}=\frac{2\sqrt{3}}{3}\arctan(t)+C$$
Since $x=u^2$ and $t=\frac{u}{\sqrt{3}}$ we have $t=\frac{\sqrt{x}}{\sqrt{3}}=\sqrt{\frac{x}{3}}\implies$ $$=\frac{2\sqrt{3}}{3}\arctan\left(\sqrt{\frac{x}{3}}\right)+C$$
