calculus - 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?

Answer

Your second question is indeed, despite your doubts, about the substitution method. The reason you don't see a $du$ (or any other variable change) is because they did it in a less-common way notationally, but if you try it with substitution you will see that it all works out the same.

The first question is a good one, because it is often not enough to know that something works, but instead you need to know how to do similar things in the future. The solution writer noticed that the integral was difficult as written, but that they could use substitution (as mentioned above) if they had two rational expressions, each containing $x$ in just one place. Since the $4$ was under the square root, they went ahead and tried adding and substracting that from the numerator to get the square of the denominator, cancelling it, on one rational, and just a constant on the other. It worked in this case. It won't always work.

Try the same logic with the integrands below to see how it goes:

$$\dfrac{2x}{5-2x}$$

$$\dfrac{2x}{5-3x}$$

Perhaps after trying both of those you can even conjecture about when the method will work and when it won't!