What is the problem with this method while integrating $(e^x-(2x+3)^4)^3$?
I already know what is its integration. I collected the answers from Quora (black ones) and WolframAlpha website (the red one).
Alright, then I tried to solve the integration in this method. But, the answer appeared different. The mathematical approach seemed very legitimate to me. Where did I go wrong, would you kindly point that out?
here's my answer
now here is the difference between my itegrals(1) graph and the graph that gave wolphrapalpha(2).
I used same method to solve this two functions i found. and it worked.
BUT WHY IT DID NOT WORK FOR THAT ONE, WHERE DID I WENT WRONG?
Your methodology is highly nonstandard and suspicious, but I will try to formalize it.
You appear to deal with integrals of the form $$ \int f(x)^n dx $$
I think your general strategy is
Set $y= f(x)$
Integrate $y^n$ to get $y^{n+1}/(n+1)$.
Divide $y^{n+1}/(n+1)$ by $dy$ and then write everything in terms of $x$.
So in the end you are saying that the following is an antiderivative for $f(x)^n$ $$ \frac{f(x)^{n+1}}{(n+1) f'(x)}\tag{1} $$
So let's take the derivative of the last function and see what we get (we should get $f(x)^n$ if your strategy is correct). Applying the quotient rule and chain rule, we get the derivative of the function in $(1)$ as:
$$ \frac{f(x)^n (f'(x))^2-\frac{1}{n+1}f(x)^{n+1} f''(x)}{(f'(x))^2}\tag{2} $$
In general, there is no reason to expect that the function in $(2)$ is going to equal $f(x)^n$.
On the other hand, if we are in the special case that $f'(x)$ is a nonzero constant, then the expression in $(2)$ is equal to $f(x)^n$ since in this case $f''(x)=0$. This is precisely what happens in your two easier examples where the strategy works. But this is not the case in the more complicated example.