Compute $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$ 
Evaluate $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$

I tried so many substitutions but none of them led me to the right answer:
$u=\frac 1{\sqrt{x+3}}$, $u=\frac 1{x+3}$, $u=\sqrt{x}$... I even got to something like $\int_0^1 \frac {u^2}{(u^2+3)^{\frac 32}}du$ or $\int_0^1 \frac {\sqrt{1-3u^2}}{u}du$... and I don't know how to solve these...

If you change $u=\sqrt{x} \Rightarrow u^2=x \Rightarrow 2udu=dx$, then:
$$\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx=\int_0^1\frac {2u^2}{(u^2+3)^{3/2}}du=\int_0^1\frac {2u^2+6-6}{(u^2+3)^{3/2}}du=\\
2\int_0^1\frac {1}{(u^2+3)^{1/2}}du-6\int_0^1\frac {1}{(u^2+3)^{3/2}}du.$$
Both integrals you can evaluate by $u=\sqrt{3}\tan t$. See this and this.
