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Why does the formula \[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]give the wrong result if we put \(n=-1\) ?

Calculus1
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AS you only stated the formula, substituting n = -1 in the formula on RHS, gives undefined form
This not the answer I want to hear. Why does this formula give the right result if we put any \(n\) except \(n=-1\) ? May be there is any trick for \(n=-1\) ?
Because the antiderivative of x^{-1} is ln(x)

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Other answers:

\[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C\]What is wrong with it?
Simply, 1/0 does not exist :)
\[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C=\ln x+C\]\[\frac10=\ln x\]
Maybe it's best to recall that we have this formula \[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]BECAUSE of this formula \[\Large \frac{d}{dx}x^n=nx^{n-1}\]
Or, more relevantly \[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]
But you see that it no longer applies for n = -1
Because \[\Large \frac{d}{dx}\frac{x^{-1+1}}{-1+1}=\frac{d}{dx}\frac10\]and 1/0 never existed to begin with :/
Ok. It is time to tell you the truth.
I do love the truth :)
Noting of course, that there are many ways to explain this.
Try to compute\[\frac{d}{dx}\left(\frac{x^{n+1}-1}{n+1}\right)\]
Well, of course, it's still \[x^n\]
But now if you put \(n=-1\) what will you get?
Indeterminate.
Don't mind on indeterminate. Just compute this like a simple limit.
0/0
\(\frac{\sin n}{n}\) is \(\frac00\) too, but we all know that it is equal to 1. What is that limit?
I really don't know.
Do you know the L'hopital's rule?
No...
\[\lim_{n\rightarrow-1}\frac{x^{n+1}-1}{n+1}\]Anybody here know how to compute this limit? @ParthKohli
@PeterPan Do you know what is Wolfram Alpha?
That's why Google is here. :)
It seems its limit is ln(x)
Yes.
Your point?
Sorry, cant understand what means "Your point?".
Never mind. :)
You did say you were looking for a specific answer, so there was preference :P
Besides, you gave a specific example, which is not valid for proofs :) It has to be true for all constants. I think I'll stand by my original statements, interesting though this was...
Actually, the right expression is \[\int x^n\,dx=\frac{x^{n+1}-1}{n+1}+C\]
Now we can include \(n=-1\) also.
No you can't. It has to be \[\Large \int x^k\,dx=\lim_{k\rightarrow n}\frac{x^{n+1}-1}{n+1}+C\]
\[\Large \int x^k\,dx=\lim_{n\rightarrow k}\frac{x^{n+1}-1}{n+1}+C\]
Without the limit, it still won't make sense.
Very nice. But forget using limits. It is restrictions that really does not play a big role.
Finally. But I find it easier to just affirm that the modified power rule \[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]does not hold for n=-1 And just proving that the derivative of ln(x) is 1/x,
the diff is there since 1/x is not continuous at x=0 x^n always passes through (0,0) if it is not equal to -1, but for n=-1, the shape of graph is completely different. And, integration is simply the area under the graph. I know my logic is very wayward and not very strong, but still.. a point worth mentioning. :P
^^

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