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.
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 $2x+3=3\sec y$
Let $x=3\tan^2\theta$, so that $x+3=3\sec^2\theta$ and $dx=6\tan\theta\sec^2\theta\,d\theta$ with $x=0\implies\theta=0$ and $x=1\implies\tan\theta=1/\sqrt3\implies\theta=\pi/6$, and thus
$$\int_0^1{\sqrt x\over(x+3)\sqrt{x+3}}dx=\int_0^{\pi/6}{\sqrt3\tan\theta\over3\sqrt3\sec^3\theta}6\tan\theta\sec^2\theta\,d\theta=2\int_0^{\pi/6}{\tan^2\theta\over\sec\theta}d\theta\\ =2\int_0^{\pi/6}(\sec\theta-\cos\theta)d\theta\\ =2\left(\ln|\sec\theta+\tan\theta|-\sin\theta \right)\Big|_0^{\pi/6}\\ =2(\ln|\sec(\pi/6)+\tan(\pi/6)|-\sin(\pi/6))\\ =2\left(\ln\left({2\over\sqrt3}+{1\over\sqrt3} \right)-{1\over2} \right)\\ =\ln3-1\approx0.0986$$
They key idea here was to get rid of all the square roots, changing things into a purely rational trig integral. We were lucky in that the limits of integration changed nicely; if they hadn't, the answer would have been a bit messy.
