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 
 A: 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 $t=\frac{\sqrt{x}}{\sqrt{3}}=\sqrt{\frac{x}{3}}\implies$ $$=\frac{2\sqrt{3}}{3}\arctan\left(\sqrt{\frac{x}{3}}\right)+C$$
A: 
Problem
$$ \int \frac{\mathrm{d}x}{\sqrt{x}(x + 3)} $$

Solution: Substitute $\displaystyle u = \frac{\sqrt{x}}{\sqrt{3}} \rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{2\sqrt{3}\sqrt{x}}$ Which, means $\mathrm{d}x = 2 \sqrt{3}\sqrt{x}\,\mathrm{d}u$.
$$ \int \frac{\mathrm{d}x}{\sqrt{x}(x + 3)} = \int \frac{2\sqrt{3}}{3 u^2 + 3}\,\mathrm{d}u = \frac{2}{\sqrt{3}} \int \frac{\mathrm{d}u}{1 + u^2}$$
Where I leave the last step as an exercise to the reader. 

Contrast this answer to the one made by upanddownintegrate and you will see that by cleverly selecting the constant a in the substitution $u = a \sqrt{x}$ we simplified the last step. However, in practice we usually just use two substitutions in stead, as it is more time consuming to find the most clever substitution
$$
\int \frac{\mathrm{d}x}{\sqrt{x}(x+3)} 
\stackrel{u \mapsto \sqrt{x}}{=}
2 \int \frac{\mathrm{d}u}{u^2 + 3} 
\stackrel{y \mapsto \sqrt{3}u}{=}
\frac{2}{\sqrt{3}}\int \frac{\mathrm{d}y}{1 + y^2} 
$$

I will leave it to you to use $u \mapsto a \sqrt{x}$ on the integral, and then factorize it to see why $a = 1/\sqrt{3}$ is a good value.
A: If you've set $u=\sqrt x/\sqrt 3,$ then we have that $u\sqrt 3=\sqrt x,$ so that $3u^2=x.$ Therefore, we have that $\mathrm dx=6u\mathrm du.$ You can now make the appropriate substitutions for $\sqrt x,\,x$ and $\mathrm dx$ in the integral.
A: I think the difficulty in this is understanding where the magic of certain substitutions come from. To tackle this, let's investigate a simpler problem first. I will type out the thought process in all its gory detail as this will let us see exactly what is going on. Consider 
$$ \int \frac{2}{2x+3} {\rm d}x $$
Obviously we could make the substitution $u=2x+3$ immediately. However, if we were to not have that leap of insight, we could choose to substitute $y=2x$ first, which will give us
$$ \int \frac{1}{y+3} {\rm d}y$$
Then we possibly try to make another substitution of $u=y+3$, resulting with
$$\int \frac{1}{u} {\rm d}u = \ln|u| +c$$
Now, if we were to retrace our substitution journey, we would find that $y+3=u = 2x +3$. And here we witness the fact that we could have made the substitution in just one move from the beginning! This may not seem that magical, since this example was so easy.. but let's look at another example.
$$ \int_1^\sqrt{2} \frac{1}{(1+y^2)\sqrt{2-y^2}} {\rm d}y $$
We are thrown this monster of an integral and have no clue where to begin, so maybe we mess around and decide make our first substitution $u=1/y$ (note: this is a common substitution to attempt given an algebraic function as the integrand). We get
$$\int_\frac11^\frac1{\sqrt2} \frac{1}{(1+\frac1{u^2})\sqrt{2-\frac1{u^2}}} \left( -\frac1{u^2} \right) {\rm d}u 
\\ = \int_\frac1{\sqrt2}^1 \frac{u}{(u^2+1)\sqrt{2u^2-1}} {\rm d}u$$
This then begs us to make the substitution $x=u^2$, but we could be smarter. If we were to choose $x=u^2+1$, that would skip the extra step of tedious working while still dealing with that pestering $u$ on the top. However, this simplifies the terms inside the brackets, and we would more likely rather to get rid of the terms in the square root since those tend to be much harder to deal with. So, we instead make the substitution $x=2u^2-1$. We now get
$$\int_0^1 \frac{\frac14}{\left( \frac12 (x+1) +1 \right) \sqrt{x}} {\rm d}x
\\ = \frac12 \int_0^1 \frac1{(x+3)\sqrt{x}} {\rm d}x$$
Which looks awfully familiar...
And indeed, if we were to try and compress the long train of substitutions: $u=1/y$, $x=2u^2-1$ and finally $t=\frac{\sqrt{x}}{\sqrt3}$:
$$t=\frac{\sqrt{x}}{\sqrt3} = \frac{\sqrt{2u^2 - 1}}{\sqrt3} 
= \frac{\sqrt{\frac2{y^2} - 1}}{\sqrt3}
= \frac{\sqrt{2-y^2}}{\sqrt3 y}$$
We would get the correct answer of $\frac{\pi\sqrt3}{18}$ with the immediate substitution of $t=\frac{\sqrt{2-y^2}}{\sqrt3 y}$ in the case of my example.
