Evaluating $\int_0^1 \frac{3x}{\sqrt{4-3x}} dx$ 
So this is the integral I must evaluate:
$$\int_0^1 \frac{3x}{\sqrt{4-3x}} dx$$

I have this already evaluated but I don't understand one of the steps in its transformation.
I understand how integrals are evaluated, but I don't understand some of the steps when it is being broken down and integrated.
The steps are as follows:
$$ - \left( \frac {-3x} {\sqrt{4-3x}}\right)$$
$$ - \left( \frac {4-3x-4}{\sqrt{4-3x}}\right) $$
$$ - \sqrt {4-3x} + \frac {4}{\sqrt{4-3x}}$$
$\mathbf {Question1} $ In these three steps the first thing I don't understand how it got broken down into two terms in the third step. If I add together the terms of the third step I get back the original one but I don't understnad how the author reached this point in the first place, like how can I know how to break it down into which and which terms?
After this it is put back into the original equation:
$$ \int_0^1 \frac {3x}{\sqrt{4-3x}} dx = - \int_0^1 (4-3x)^\frac {1}{2} dx + 4 \int_0^1 (4-3x)^\frac{-1}{2} dx $$
$$ =  \frac {1}{3} \int_0^1 (4-3x)^\frac {1}{2} (-3 dx) - \frac{4}{3} \int_0^1 (4-3x)^\frac{-1}{2} (-3dx) $$
After this it is integrated as usual with $-3dx$ term disappearing in both and $(4-3x)^\frac{1}{2}$ and $(4-3x)^\frac{-1}{2}$ get integrated with n+1 formula.
$\mathbf{Question 2}$ Why was $-3$ multiplied and divided in second step. The dx doesn't change to du so it clearly isn't substitution. So what exactly is happening here?

Addressing question 1:
\begin{align*}
& - \left( \frac {4-3x-4}{\sqrt{4-3x}}\right) \\
= & - \left( \frac {4-3x}{\sqrt{4-3x}} - \frac{4}{\sqrt{4 - 3x}}\right) \\
= & - \frac {(\sqrt{4-3x})^2}{\sqrt{4-3x}} - \left(- \frac{4}{\sqrt{4 - 3x}}\right) \\
= & - \sqrt {4-3x} + \frac {4}{\sqrt{4-3x}}
\end{align*}
