Mathematics Asked on January 3, 2022

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.

Answered by halrankard on January 3, 2022

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