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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).

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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

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now here is the difference between my itegrals(1) graph and the graph that gave wolphrapalpha(2).

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I used same method to solve this two functions i found. and it worked.

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BUT WHY IT DID NOT WORK FOR THAT ONE, WHERE DID I WENT WRONG?
halrankard
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AL vees
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  • How do you get $$f(z) \ dz= \frac{z^4}{4}$$ from $$f(z)=z^3$$?? – Vishu Jul 24 '20 at 18:56
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    It seems like you are doing some integration/substitution steps out of order. Remember that you can't find the antiderivative until you have completed the substitution. – DMcMor Jul 24 '20 at 18:58
  • @DMcMor where the substitution needs to be completed? would you mind showing me that? – AL vees Jul 25 '20 at 03:46

1 Answers1

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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

  1. Set $y= f(x)$

  2. Integrate $y^n$ to get $y^{n+1}/(n+1)$.

  3. 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.

halrankard
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  • Thank you very much! Also, would you tell me where to improve the method? So that we can solve complex ones? I have a feeling that the mathematical structure of the method is ok, but there is a problem inside, in math manipulation – AL vees Jul 25 '20 at 03:56
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    The mathematical structure of the method is $u$-substitution, and so the way to improve the method is to do it properly to see if it actually works. You are attempting the substitution $u=e^x+(2x+3)^4$, which means $$dx=\frac{du}{e^x+8(2x+3)^3}$$. Then $$\int (e^x+(2x+3)^4)^3 dx=\int \frac{u^4}{e^x+8(2x+3)^3} du$$ In order successfully implement the method you need to completely substitute all of this to get something in terms of $u$. I'll be honest that it doesn't look too promising, and you're probably going to end up with a new integral that is even more complicated. – halrankard Jul 25 '20 at 12:56
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    So this means one of two things, either you need to pick a better choice of $u$, or $u$-substitution is not going to work. My guess is the latter option. Not every integral can be done with $u$-substitution. The "brute force" way to do this particular integral is to multiply out the cube $(e^x+(2x+3)^4)^3$ and then integrate each term individually. This will require integration by parts. I'm not saying that there aren't clever simplifications lying around, but going by the answers in your screenshots, this is what the computer programs did. – halrankard Jul 25 '20 at 13:00
  • I have one important question, in which cases I can't use u substitution? – AL vees Jul 25 '20 at 14:42
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    Unfortunately your question has no better answer than "in any case where $u$-sub doesn't work". There is no simple algorithm or step-by-step instructions that tell you when to use one integration strategy over another. You just have to do a lot a practice problems to build intuition for when certain ones will work, and to learn the clues that guide you toward one strategy versus another. You can probably find reading online. For example: https://math.stackexchange.com/questions/538654/when-to-do-u-substitution-and-when-to-integrate-by-parts – halrankard Jul 25 '20 at 15:00
  • is there any common procedure? – AL vees Jul 25 '20 at 15:03
  • i have also found that if you add −f(x)n+1(n+1)f′(x)+1f′(x)n to the equation no-1 that you gave in the answer; the derivative turns out to be f(x)^n – AL vees Jul 25 '20 at 15:06
  • −f(x)^(n+1)/(n+1)f′(x)+1/f′(x)^n sorry, this the right one – AL vees Jul 25 '20 at 15:09
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    It is difficult to tell what are the numerators and the denominators in what your wrote. But it looks to me like if you add that to (1) you will just get $$\frac{1}{f'(x)^n}$$ whose derivative is certainly not $f(x)^n$. But anyway, we should avoid this back and forth. If you have a new question about a proposed function whose derivative will always be $f(x)^n$ then you can ask it in a new question. But please use latex to format the mathematics so that it is clear to read. – halrankard Jul 25 '20 at 15:14