<s>[INST] How to calculate this integral without any integration techniques? 
Differentiate  $f(x) = (5x+2)\ln(2x+1)$ with respect to $x$. Hence, find $\int \ln(2x+1)^3dx$.

Because of the word "Hence" I'm assuming that the question doesn't allow integration techniques such as integration by parts or substitution.
The first part is trivial. The derivative is $5\ln(2x+1) + \dfrac{2(5x+2)}{2x+1}$. Now my line of thought so far has been to somehow get this derivative to the desired result: $\ln(2x+1)^3$ and then use $f(x)$ to get the integral. However I don't see any straight way to do this.
So I decided to experiment a little. Firstly I differentiated $(3x+2)\ln(2x+1)$ and got $$3\ln(2x+1) + \dfrac{2(3x+2)}{2x+1}$$ Now at least I have the $\ln(2x+1)^3$ term but I have another complication: $\dfrac{2(3x+2)}{2x+1}$.  
I decided to take a look at the answer: $3x\ln(2x+1) + \dfrac{3}{2}\ln(2x+1) - 3x + C$. I thought that if I differentiated this, I would have some idea of where I should go, and how to proceed. So that's what I did, and I got:
$$3\ln(2x+1) + \dfrac{2(3x)}{2x+1} + \dfrac{3}{2x+1} - 3$$ This is perfect because $\dfrac{2(3x)}{2x+1} + \dfrac{3}{2x+1} - 3$ cancels out perfectly. But I have no idea how to get here, starting from $f'(x)$. But I'd say that my first experiment was pretty close. Any help would be highly appreciated.
 [/INST] Alright so from my first experiment we have the derivatie $3\ln(2x+1) + \dfrac{2(3x+2)}{2x+1}$. To get this to just $3\ln(2x+1)$ we subtract $3$ and $\dfrac{1}{2x+1}$. This gives us the expression $$\bigg(3\ln(2x+1) + \dfrac{6x}{2x+1} + \dfrac{4}{2x+1} \bigg) - 3 - \dfrac{1}{2x+1}$$ Integrating gives us $(3x+2)\ln(2x+1) - 3x  - \dfrac{1}{2}\ln(2x+1) + C$ 
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