<s>[INST] 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...
 [/INST] Hint:
Integrate by parts
$$\int\sqrt x\dfrac1{(x+3)^{3/2}}dx$$
$$=\sqrt x\int\dfrac1{(x+3)^{3/2}}dx-\int\left(\dfrac{d(\sqrt x)}{dx}\int\dfrac1{(x+3)^{3/2}}dx\right)dx=?$$
Now for $\displaystyle\int\dfrac{dx}{\sqrt{x(x+3)}}$
$x(x+3)=\dfrac{(2x+3)^2-3^2}4$
set $2x+3=3\sec y$
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